C语言逆向分析——Switch语句，为何大多数情况较if语句更高效？就是因为查找表

Switch语句

Switch语句也是分支语句的一种，其语法如下：

switch(表达式)
{
 case 常量表达式1:
 语句;
 break;
 case 常量表达式:
 语句;
 break;
 case 常量表达式:
 语句;
 break;
 ......
 default:
 语句;
 break;
}

需要注意如下几点：

表达式结束不能是浮点数
case后的常量值不能一样
case后的值必须是常量
break非常重要，当执行到一个分支后，如果没有break就会继续向下执行，遇到break才会跳出switch语句
default语句与位置无关，但是当default写在其他条件的前面时。如果没有break就会向下继续匹配执行

switch语句与if..else语句的区别：

switch语句只能进行等值判断（仅能支持常量表达式），而if..else可以进行区间判断（表达式、常量...都可以）
switch语句的执行效率远远高于if..else，在分支条件比较多的情况下，这种趋势愈发明显

Switch语句为什么高效

之前说到switch语句的执行效率远远高于if..else，这是为什么？我们可以写一段代码通过反汇编来查看其在底层的具体实现：

int x = 3;
 
switch(x)
{
 case 1:
 printf("A \n");
 break;
 case 2:
 printf("B \n");
 break;
 case 3:
 printf("C \n");
 break;
 case 4:
 printf("D \n");
 break;
 default:
 printf("default \n");
 break;
}

其反汇编代码为：

images/download/attachments/12714021/image2021-2-22_0-40-17.png

如上反汇编代码我们可以看出switch在一开始就直接将变量x-1与3进行比较，ja指令则表示大于则跳转（a表示above），但这变量x明显等于3，所以将值赋予edx，最后jmp跳到堆栈地址2*4+40d896（40d896+8），也就是0x40D89E对应的值：

images/download/attachments/12714021/image2021-2-22_0-51-14.png

最终跳转到0x0040d80c，打印出了C。

我自己机器上的实验：

为啥本质上是查找表呢？我自己机器上分析下：

那么我们再使用if...else来实现相同功能：

int x = 3;
 
if (x == 1) {
 printf("A \n");
} else if (x == 2) {
 printf("B \n");
} else if (x == 3) {
 printf("C \n");
} else if (x == 4) {
 printf("D \n");
} else {
 printf("default \n");
}

其反汇编代码为：

images/download/attachments/12714021/image2021-2-22_0-43-25.png

可以看见其会一行一行的执行，而不像switch语句一样，直接可以跳转到对应的地址。

通过观察堆栈，我们可以发现switch的高明之处：

images/download/attachments/12714021/image2021-2-22_0-53-37.png

其将case分支的执行地址都存入到了堆栈中（也就是查找表），而后使用算法去寻找到对应要执行的地址（堆栈存储的地址），这套算法我们可以这样理解：

1.堆栈中会存在一个执行地址的内存表，其顺序是按照case后的常量大小排序的，最大的常量所在的分支执行地址最先压入表中；如下图所示我将代码顺序打乱后还是如此：

images/download/attachments/12714021/image2021-2-22_1-14-30.png

2.根据内存表存储顺序，直接将传入的值减去1，然后*4（这里也是数据宽度4字节）+最后压入的堆栈地址。

了解完算法原理之后，我们需要了解一个概念：大表和小表；当生成的函数跳转地址表，每个成员有4个字节的时，我们称之为大表。

注意：

在case分支小于等于3个、case最大值和最小值差大于等于255时不存在大小表
在case分支大于3个并且大表项空隙小于等于6个时只有大表没有小表
在case分支大于3个且大表项空隙大于6个且case最大最小差值小于255时存在大表和小表

最后：在一般情况下，我们通常会去使用if语句而不是switch，因为if语句在编程时更加得心应手，所以我们可以暂且忽视效率而去选择更加便利的编写方式。

实例

case分支少的switch语句

CPP代码：

#include "stdafx.h"
void Fun(int x) {
	switch (x) {
		case 1:
			printf("%d",1);
			break;
		case 2:
			printf("%d",2);
			break;
		case 3:
			printf("%d",3);
			break;
		default:
			printf("None");
			break;
	}
}

int main(int argc, char* argv[]) {
	Fun(2);
	return 0;
}

反汇编：

Fun:
0040D490   push        ebp
0040D491   mov         ebp,esp
0040D493   sub         esp,44h
0040D496   push        ebx
0040D497   push        esi
0040D498   push        edi
0040D499   lea         edi,[ebp-44h]
0040D49C   mov         ecx,11h
0040D4A1   mov         eax,0CCCCCCCCh
0040D4A6   rep stos    dword ptr [edi]
0040D4A8   mov         eax,dword ptr [ebp+8]
0040D4AB   mov         dword ptr [ebp-4],eax
0040D4AE   cmp         dword ptr [ebp-4],1
0040D4B2   je          Fun+32h (0040d4c2)
0040D4B4   cmp         dword ptr [ebp-4],2
0040D4B8   je          Fun+43h (0040d4d3)
0040D4BA   cmp         dword ptr [ebp-4],3
0040D4BE   je          Fun+54h (0040d4e4)
0040D4C0   jmp         Fun+65h (0040d4f5)
0040D4C2   push        1
0040D4C4   push        offset string "%d" (0042210c)
0040D4C9   call        printf (0040d760)
0040D4CE   add         esp,8
0040D4D1   jmp         Fun+72h (0040d502)
0040D4D3   push        2
0040D4D5   push        offset string "%d" (0042210c)
0040D4DA   call        printf (0040d760)
0040D4DF   add         esp,8
0040D4E2   jmp         Fun+72h (0040d502)
0040D4E4   push        3
0040D4E6   push        offset string "%d" (0042210c)
0040D4EB   call        printf (0040d760)
0040D4F0   add         esp,8
0040D4F3   jmp         Fun+72h (0040d502)
0040D4F5   push        offset string "None" (00422f6c)
0040D4FA   call        printf (0040d760)
0040D4FF   add         esp,4
0040D502   pop         edi
0040D503   pop         esi
0040D504   pop         ebx
0040D505   add         esp,44h
0040D508   cmp         ebp,esp
0040D50A   call        __chkesp (004010a0)
0040D50F   mov         esp,ebp
0040D511   pop         ebp
0040D512   ret

小结：

case分支小于4个时，其反汇编与if...else类似

case分支多的switch语句

CPP代码：

#include "stdafx.h"
void Fun(int x) {
	switch (x) {
		case 1:
			printf("1");
			break;
		case 2:
			printf("2");
			break;
		case 3:
			printf("3");
			break;
		case 4:
			printf("4");
			break;
		default:
			printf("None");
			break;
	}
}

int main(int argc, char* argv[]) {
	Fun(2);
	return 0;
}

反汇编：

Fun:
0040D490   push        ebp
0040D491   mov         ebp,esp
0040D493   sub         esp,44h
0040D496   push        ebx
0040D497   push        esi
0040D498   push        edi
0040D499   lea         edi,[ebp-44h]
0040D49C   mov         ecx,11h
0040D4A1   mov         eax,0CCCCCCCCh
0040D4A6   rep stos    dword ptr [edi]
0040D4A8   mov         eax,dword ptr [ebp+8]
0040D4AB   mov         dword ptr [ebp-4],eax
0040D4AE   mov         ecx,dword ptr [ebp-4]
0040D4B1   sub         ecx,1
0040D4B4   mov         dword ptr [ebp-4],ecx	;[ebp-4]=[ebp+8]-1
0040D4B7   cmp         dword ptr [ebp-4],3
0040D4BB   ja          $L537+11h (0040d50b)		;如果[ebp-4]>3跳转到0040d50b
0040D4BD   mov         edx,dword ptr [ebp-4]
0040D4C0   jmp         dword ptr [edx*4+40D529h]
$L533:
0040D4C7   push        offset string "1" (00422fac)
0040D4CC   call        printf (0040d760)
0040D4D1   add         esp,4
0040D4D4   jmp         $L539+1Ch (0040d510)
$L535:
0040D4D6   push        offset string "2" (00422fa8)
0040D4DB   call        printf (0040d760)
0040D4E0   add         esp,4
0040D4E3   jmp         $L539+1Ch (0040d510)
$L537:
0040D4E5   push        offset string "3" (00422fa4)
0040D4EA   call        printf (0040d760)
0040D4EF   add         esp,4
0040D4F2   jmp         $L539+1Ch (0040d510)
$L539:
0040D4F4   push        offset string "4" (0042210c)
0040D4F9   call        printf (0040d760)
0040D4FE   add         esp,4
0040D501   jmp         $L539+1Ch (0040d510)
0040D503   push        offset string "None" (00422f6c)
0040D508   call        printf (0040d760)
0040D50D   add         esp,4
0040D510   pop         edi
0040D511   pop         esi
0040D512   pop         ebx
0040D513   add         esp,44h
0040D516   cmp         ebp,esp
0040D518   call        __chkesp (004010a0)
0040D51D   mov         esp,ebp
0040D51F   pop         ebp
0040D520   ret

 EAX = 00000002 EBX = 7FFDE000
 ECX = 00000001 EDX = 00000001
 ESI = 00000000 EDI = 0012FF28
 EIP = 0040D4C0 ESP = 0012FED8
 EBP = 0012FF28 EFL = 00000293

Memory：

0040D521  C7 D4 40 00  窃@.
0040D525  D6 D4 40 00  衷@.
0040D529  E5 D4 40 00  逶@.
0040D52D  F4 D4 40 00  粼@.

小结：

当case分支大于等于4个且连续时，会在一段连续内存中存储每个case所对应语句块的起始地址（称其为大表），根据[edx*4+40D529h]即可确定跳转地址

case后的值打乱顺序

CPP代码：

#include "stdafx.h"
void Fun(int x) {
	switch (x) {
		case 2:
			printf("2");
			break;
		case 1:
			printf("1");
			break;
		case 4:
			printf("4");
			break;
		case 3:
			printf("3");
			break;
		default:
			printf("None");
			break;
	}
}

int main(int argc, char* argv[]) {
	Fun(2);
	return 0;
}

反汇编：

Fun:
0040D490   push        ebp
0040D491   mov         ebp,esp
0040D493   sub         esp,44h
0040D496   push        ebx
0040D497   push        esi
0040D498   push        edi
0040D499   lea         edi,[ebp-44h]
0040D49C   mov         ecx,11h
0040D4A1   mov         eax,0CCCCCCCCh
0040D4A6   rep stos    dword ptr [edi]
0040D4A8   mov         eax,dword ptr [ebp+8]
0040D4AB   mov         dword ptr [ebp-4],eax
0040D4AE   mov         ecx,dword ptr [ebp-4]
0040D4B1   sub         ecx,1
0040D4B4   mov         dword ptr [ebp-4],ecx
0040D4B7   cmp         dword ptr [ebp-4],3
0040D4BB   ja          $L539+0Fh (0040d503)
0040D4BD   mov         edx,dword ptr [ebp-4]
0040D4C0   jmp         dword ptr [edx*4+40D521h]
$L533:
0040D4C7   push        offset string "2" (00422fac)
0040D4CC   call        printf (0040d760)
0040D4D1   add         esp,4
0040D4D4   jmp         $L539+1Ch (0040d510)
$L535:
0040D4D6   push        offset string "1" (00422fa8)
0040D4DB   call        printf (0040d760)
0040D4E0   add         esp,4
0040D4E3   jmp         $L539+1Ch (0040d510)
$L537:
0040D4E5   push        offset string "4" (00422fa4)
0040D4EA   call        printf (0040d760)
0040D4EF   add         esp,4
0040D4F2   jmp         $L539+1Ch (0040d510)
$L539:
0040D4F4   push        offset string "3" (0042210c)
0040D4F9   call        printf (0040d760)
0040D4FE   add         esp,4
0040D501   jmp         $L539+1Ch (0040d510)
0040D503   push        offset string "None" (00422f6c)
0040D508   call        printf (0040d760)
0040D50D   add         esp,4
0040D510   pop         edi
0040D511   pop         esi
0040D512   pop         ebx
0040D513   add         esp,44h
0040D516   cmp         ebp,esp
0040D518   call        __chkesp (004010a0)
0040D51D   mov         esp,ebp
0040D51F   pop         ebp
0040D520   ret

 EAX = 00000002 EBX = 7FFDB000
 ECX = 00000001 EDX = 00000001
 ESI = 00000000 EDI = 0012FF28
 EIP = 0040D4C0 ESP = 0012FED8
 EBP = 0012FF28 EFL = 00000293

Memory：

0040D521  D6 D4 40 00  衷@.
0040D525  C7 D4 40 00  窃@.
0040D529  F4 D4 40 00  粼@.
0040D52D  E5 D4 40 00  逶@.

小结：

在值连续的情况下，顺序并不会影响生成大表

起始值对反汇编的影响

CPP代码：

#include "stdafx.h"
void Fun(int x) {
	switch (x) {
		case 100:
			printf("100");
			break;
		case 101:
			printf("101");
			break;
		case 102:
			printf("102");
			break;
		case 103:
			printf("103");
			break;
		case 104:
			printf("104");
			break;
		case 105:
			printf("105");
			break;
		case 106:
			printf("106");
			break;
		case 107:
			printf("107");
			break;
		case 108:
			printf("108");
			break;
		case 109:
			printf("109");
			break;
		default:
			printf("None");
			break;
	}
}

int main(int argc, char* argv[]) {
	Fun(102);
	return 0;
}

反汇编：

Fun:
0040D7E0   push        ebp
0040D7E1   mov         ebp,esp
0040D7E3   sub         esp,44h
0040D7E6   push        ebx
0040D7E7   push        esi
0040D7E8   push        edi
0040D7E9   lea         edi,[ebp-44h]
0040D7EC   mov         ecx,11h
0040D7F1   mov         eax,0CCCCCCCCh
0040D7F6   rep stos    dword ptr [edi]
0040D7F8   mov         eax,dword ptr [ebp+8]
0040D7FB   mov         dword ptr [ebp-4],eax
0040D7FE   mov         ecx,dword ptr [ebp-4]
0040D801   sub         ecx,64h	;64h=100
0040D804   mov         dword ptr [ebp-4],ecx
0040D807   cmp         dword ptr [ebp-4],9
0040D80B   ja          $L551+0Fh (0040d8b7)
0040D811   mov         edx,dword ptr [ebp-4]
0040D814   jmp         dword ptr [edx*4+40D8D5h]
$L533:
0040D81B   push        offset string "100" (00422fc4)
0040D820   call        printf (0040d760)
0040D825   add         esp,4
0040D828   jmp         $L551+1Ch (0040d8c4)
$L535:
0040D82D   push        offset string "101" (00422fc0)
0040D832   call        printf (0040d760)
0040D837   add         esp,4
0040D83A   jmp         $L551+1Ch (0040d8c4)
$L537:
0040D83F   push        offset string "102" (00422fbc)
0040D844   call        printf (0040d760)
0040D849   add         esp,4
0040D84C   jmp         $L551+1Ch (0040d8c4)
$L539:
0040D84E   push        offset string "103" (00422fb8)
0040D853   call        printf (0040d760)
0040D858   add         esp,4
0040D85B   jmp         $L551+1Ch (0040d8c4)
$L541:
0040D85D   push        offset string "104" (00422fb4)
0040D862   call        printf (0040d760)
0040D867   add         esp,4
0040D86A   jmp         $L551+1Ch (0040d8c4)
$L543:
0040D86C   push        offset string "105" (00422fb0)
0040D871   call        printf (0040d760)
0040D876   add         esp,4
0040D879   jmp         $L551+1Ch (0040d8c4)
$L545:
0040D87B   push        offset string "106" (00422fac)
0040D880   call        printf (0040d760)
0040D885   add         esp,4
0040D888   jmp         $L551+1Ch (0040d8c4)
$L547:
0040D88A   push        offset string "107" (00422fa8)
0040D88F   call        printf (0040d760)
0040D894   add         esp,4
0040D897   jmp         $L551+1Ch (0040d8c4)
$L549:
0040D899   push        offset string "108" (00422fa4)
0040D89E   call        printf (0040d760)
0040D8A3   add         esp,4
0040D8A6   jmp         $L551+1Ch (0040d8c4)
$L551:
0040D8A8   push        offset string "109" (0042210c)
0040D8AD   call        printf (0040d760)
0040D8B2   add         esp,4
0040D8B5   jmp         $L551+1Ch (0040d8c4)
0040D8B7   push        offset string "None" (00422f6c)
0040D8BC   call        printf (0040d760)
0040D8C1   add         esp,4
0040D8C4   pop         edi
0040D8C5   pop         esi
0040D8C6   pop         ebx
0040D8C7   add         esp,44h
0040D8CA   cmp         ebp,esp
0040D8CC   call        __chkesp (004010a0)
0040D8D1   mov         esp,ebp
0040D8D3   pop         ebp
0040D8D4   ret

Memory：

0040D8D5  1B D8 40 00  .谸.
0040D8D9  2D D8 40 00  -谸.
0040D8DD  3F D8 40 00  ?谸.
0040D8E1  4E D8 40 00  N谸.
0040D8E5  5D D8 40 00  ]谸.
0040D8E9  6C D8 40 00  l谸.
0040D8ED  7B D8 40 00  {谸.
0040D8F1  8A D8 40 00  娯@.
0040D8F5  99 D8 40 00  欂@.
0040D8F9  A8 D8 40 00  ㄘ@.

 EAX = 00000066 EBX = 7FFDB000
 ECX = 00000002 EDX = 00000002
 ESI = 00000000 EDI = 0012FF28
 EIP = 0040D814 ESP = 0012FED8
 EBP = 0012FF28 EFL = 00000297

小结：

起始值并不影响大表的生成，edx = switch表达式的值 - case整型常量起始值

连续值中抹去少项

CPP代码：

#include "stdafx.h"
void Fun(int x) {
	switch (x) {
		case 100:
			printf("100");
			break;
		case 101:
			printf("101");
			break;
		case 102:
			printf("102");
			break;
		case 103:
			printf("103");
			break;
		case 106:
			printf("106");
			break;
		case 107:
			printf("107");
			break;
		case 108:
			printf("108");
			break;
		case 109:
			printf("109");
			break;
		default:
			printf("None");
			break;
	}
}

int main(int argc, char* argv[]) {
	Fun(102);
	return 0;
}

反汇编：

Fun:
0040D7E0   push        ebp
0040D7E1   mov         ebp,esp
0040D7E3   sub         esp,44h
0040D7E6   push        ebx
0040D7E7   push        esi
0040D7E8   push        edi
0040D7E9   lea         edi,[ebp-44h]
0040D7EC   mov         ecx,11h
0040D7F1   mov         eax,0CCCCCCCCh
0040D7F6   rep stos    dword ptr [edi]
0040D7F8   mov         eax,dword ptr [ebp+8]
0040D7FB   mov         dword ptr [ebp-4],eax
0040D7FE   mov         ecx,dword ptr [ebp-4]
0040D801   sub         ecx,64h
0040D804   mov         dword ptr [ebp-4],ecx
0040D807   cmp         dword ptr [ebp-4],9
0040D80B   ja          $L547+0Fh (0040d893)
0040D811   mov         edx,dword ptr [ebp-4]
0040D814   jmp         dword ptr [edx*4+40D8B1h]
$L533:
0040D81B   push        offset string "100" (00422fbc)
0040D820   call        printf (0040d760)
0040D825   add         esp,4
0040D828   jmp         $L547+1Ch (0040d8a0)
$L535:
0040D82A   push        offset string "101" (00422fb8)
0040D82F   call        printf (0040d760)
0040D834   add         esp,4
0040D837   jmp         $L547+1Ch (0040d8a0)
$L537:
0040D839   push        offset string "102" (00422fb4)
0040D83E   call        printf (0040d760)
0040D843   add         esp,4
0040D846   jmp         $L547+1Ch (0040d8a0)
$L539:
0040D848   push        offset string "103" (00422fb0)
0040D84D   call        printf (0040d760)
0040D852   add         esp,4
0040D855   jmp         $L547+1Ch (0040d8a0)
$L541:
0040D857   push        offset string "106" (00422fac)
0040D85C   call        printf (0040d760)
0040D861   add         esp,4
0040D864   jmp         $L547+1Ch (0040d8a0)
$L543:
0040D866   push        offset string "107" (00422fa8)
0040D86B   call        printf (0040d760)
0040D870   add         esp,4
0040D873   jmp         $L547+1Ch (0040d8a0)
$L545:
0040D875   push        offset string "108" (00422fa4)
0040D87A   call        printf (0040d760)
0040D87F   add         esp,4
0040D882   jmp         $L547+1Ch (0040d8a0)
$L547:
0040D884   push        offset string "109" (0042210c)
0040D889   call        printf (0040d760)
0040D88E   add         esp,4
0040D891   jmp         $L547+1Ch (0040d8a0)
0040D893   push        offset string "None" (00422f6c)
0040D898   call        printf (0040d760)
0040D89D   add         esp,4
0040D8A0   pop         edi
0040D8A1   pop         esi
0040D8A2   pop         ebx
0040D8A3   add         esp,44h
0040D8A6   cmp         ebp,esp
0040D8A8   call        __chkesp (004010a0)
0040D8AD   mov         esp,ebp
0040D8AF   pop         ebp
0040D8B0   ret

Memory：

0040D8B1  1B D8 40 00  .谸.
0040D8B5  2A D8 40 00  *谸.
0040D8B9  39 D8 40 00  9谸.
0040D8BD  48 D8 40 00  H谸.
0040D8C1  93 D8 40 00  撠@.	;缺失的104和105都跳转到0040D893
0040D8C5  93 D8 40 00  撠@.	;而0040D893是Default的语句块地址
0040D8C9  57 D8 40 00  W谸.
0040D8CD  66 D8 40 00  f谸.
0040D8D1  75 D8 40 00  u谸.
0040D8D5  84 D8 40 00  勜@.

 EAX = 00000066 EBX = 7FFDE000
 ECX = 00000002 EDX = 00000002
 ESI = 00000000 EDI = 0012FF28
 EIP = 0040D814 ESP = 0012FED8
 EBP = 0012FF28 EFL = 00000297

小结：

空缺地址通过填充default语句块地址解决，但会造成内存浪费

连续值中抹去多项

CPP代码：

#include "stdafx.h"
void Fun(int x) {
	switch (x) {
		case 100:
			printf("100");
			break;
		case 101:
			printf("101");
			break;
		case 102:
			printf("102");
			break;
		case 106:
			printf("106");
			break;
		case 108:
			printf("108");
			break;
		case 112:
			printf("112");
			break;
		case 115:
			printf("115");
			break;
		default:
			printf("None");
			break;
	}
}

int main(int argc, char* argv[]) {
	Fun(110);
	return 0;
}

反汇编：

Fun:
0040D7E0   push        ebp
0040D7E1   mov         ebp,esp
0040D7E3   sub         esp,44h
0040D7E6   push        ebx
0040D7E7   push        esi
0040D7E8   push        edi
0040D7E9   lea         edi,[ebp-44h]
0040D7EC   mov         ecx,11h
0040D7F1   mov         eax,0CCCCCCCCh
0040D7F6   rep stos    dword ptr [edi]
0040D7F8   mov         eax,dword ptr [ebp+8]
0040D7FB   mov         dword ptr [ebp-4],eax
0040D7FE   mov         ecx,dword ptr [ebp-4]
0040D801   sub         ecx,64h
0040D804   mov         dword ptr [ebp-4],ecx
0040D807   cmp         dword ptr [ebp-4],0Fh
0040D80B   ja          $L545+0Fh (0040d888)
0040D80D   mov         eax,dword ptr [ebp-4]
0040D810   xor         edx,edx				;edx清零
0040D812   mov         dl,byte ptr  (0040d8c6)[eax]	;dl=[小表地址+eax] ==>没有看懂。。。todo
0040D818   jmp         dword ptr [edx*4+40D8A6h]
$L533:
0040D81F   push        offset string "102" (00422fb4)
0040D824   call        printf (0040d760)
0040D829   add         esp,4
0040D82C   jmp         $L545+1Ch (0040d895)
$L535:
0040D82E   push        offset string "1003\n" (00422fd8)
0040D833   call        printf (0040d760)
0040D838   add         esp,4
0040D83B   jmp         $L545+1Ch (0040d895)
$L537:
0040D83D   push        offset string "109" (00422fac)
0040D842   call        printf (0040d760)
0040D847   add         esp,4
0040D84A   jmp         $L545+1Ch (0040d895)
$L539:
0040D84C   push        offset string "1009\n" (00422fd0)
0040D851   call        printf (0040d760)
0040D856   add         esp,4
0040D859   jmp         $L545+1Ch (0040d895)
$L541:
0040D85B   push        offset string "108" (00422fa4)
0040D860   call        printf (0040d760)
0040D865   add         esp,4
0040D868   jmp         $L545+1Ch (0040d895)
$L543:
0040D86A   push        offset string "109" (00422fc8)
0040D86F   call        printf (0040d760)
0040D874   add         esp,4
0040D877   jmp         $L545+1Ch (0040d895)
$L545:
0040D879   push        offset string "115" (00422fc0)
0040D87E   call        printf (0040d760)
0040D883   add         esp,4
0040D886   jmp         $L545+1Ch (0040d895)
0040D888   push        offset string "error\n" (00422f6c)
0040D88D   call        printf (0040d760)
0040D892   add         esp,4
0040D895   pop         edi
0040D896   pop         esi
0040D897   pop         ebx
0040D898   add         esp,44h
0040D89B   cmp         ebp,esp
0040D89D   call        __chkesp (004010a0)
0040D8A2   mov         esp,ebp
0040D8A4   pop         ebp
0040D8A5   ret

Memory：

;大表
0040D8A6  1F D8 40 00  .谸.
0040D8AA  2E D8 40 00  .谸.
0040D8AE  3D D8 40 00  =谸.
0040D8B2  4C D8 40 00  L谸.
0040D8B6  5B D8 40 00  [谸.
0040D8BA  6A D8 40 00  j谸.
0040D8BE  79 D8 40 00  y谸.
0040D8C2  88 D8 40 00  堌@.
;小表
0040D8C6  00 01 02 07  ....
0040D8CA  07 07 03 07  ....
0040D8CE  04 07 07 07  ....
0040D8D2  05 07 07 06  ....

小表的解释：

当空缺值太多时内存的浪费也会变多，编译器当然知道这样不是办法，所以利用小表来解决这个问题。小表可以看作是一个智能蹦床，对于不同的玩家会给出不同的力，遇到没有付费的玩家（空缺值）直接将他抛出场外（给出参数，使其跳转到default的语句块），遇到付费玩家（存在的值）则按照他的等级给出不同的力（给出参数，使其跳转到其对应的语句块）
可以看出，在小表中所有的空缺值都是07（因为在这个样例中，当edx=7时[edx*4+40D8A6h]的地址为default语句块的地址），而存在的值的对应值从0递增。

断开一定程度的选择

在这里插入图片描述

在这里插入图片描述
关键点

mov dl,byte ptr (004010d0)[eax]

jmp dword ptr [edx4+4010BCh]
(004010d0)[eax] = [004010d0 + eax4] 在这个地址中得到一个偏移作为查询大表的edx值。==》内在原理是啥？？？todo

差值大的情况

CPP代码：

#include "stdafx.h"
void Fun(int x) {
	switch (x) {
		case 301:
			printf("301");
			break;
		case 302:
			printf("302");
			break;
		case 303:
			printf("303");
			break;
		case 304:
			printf("304");
			break;
		case 305:
			printf("305");
			break;
		case 306:
			printf("306");
			break;
		case 307:
			printf("307");
			break;
		case 308:
			printf("308");
			break;
		case 309:
			printf("309");
			break;
		case 3:
			printf("3");
			break;
		default:
			printf("None");
			break;
	}
}

int main(int argc, char* argv[]) {
	Fun(102);
	return 0;
}

反汇编：

Fun:
0040D7E0   push        ebp
0040D7E1   mov         ebp,esp
0040D7E3   sub         esp,44h
0040D7E6   push        ebx
0040D7E7   push        esi
0040D7E8   push        edi
0040D7E9   lea         edi,[ebp-44h]
0040D7EC   mov         ecx,11h
0040D7F1   mov         eax,0CCCCCCCCh
0040D7F6   rep stos    dword ptr [edi]
0040D7F8   mov         eax,dword ptr [ebp+8]
0040D7FB   mov         dword ptr [ebp-4],eax
0040D7FE   cmp         dword ptr [ebp-4],131h	;0x131 = 305
0040D805   jg          Fun+75h (0040d855)		;[ebp-4] > 305 跳转到0040d855
0040D807   cmp         dword ptr [ebp-4],131h
0040D80E   je          Fun+0D7h (0040d8b7)		;[ebp-4] == 305 跳转到0040d8b7
0040D814   cmp         dword ptr [ebp-4],12Eh	;0x12E = 302
0040D81B   jg          Fun+5Eh (0040d83e)		;[ebp-4] > 302 跳转到0040d855
0040D81D   cmp         dword ptr [ebp-4],12Eh
0040D824   je          Fun+0A7h (0040d887)		;[ebp-4] == 302 跳转到0040d887
0040D826   cmp         dword ptr [ebp-4],3;   3比较
0040D82A   je          $L549+0Fh (0040d902)
0040D830   cmp         dword ptr [ebp-4],12Dh; ==>301比较
0040D837   je          Fun+95h (0040d875)
0040D839   jmp         $L549+1Eh (0040d911)
0040D83E   cmp         dword ptr [ebp-4],12Fh
0040D845   je          Fun+0B9h (0040d899)
0040D847   cmp         dword ptr [ebp-4],130h;==>304比较
0040D84E   je          Fun+0C8h (0040d8a8)
0040D850   jmp         $L549+1Eh (0040d911)
0040D855   mov         ecx,dword ptr [ebp-4]
0040D858   sub         ecx,132h ;306比较
0040D85E   mov         dword ptr [ebp-4],ecx
0040D861   cmp         dword ptr [ebp-4],3
0040D865   ja          $L549+1Eh (0040d911)
0040D86B   mov         edx,dword ptr [ebp-4]
0040D86E   jmp         dword ptr [edx*4+40D92Fh] ;==>这里也有表
0040D875   push        offset string "301" (00422fc4)
0040D87A   call        printf (0040d760)
0040D87F   add         esp,4
0040D882   jmp         $L549+2Bh (0040d91e)
0040D887   push        offset string "302" (00422fc0)
0040D88C   call        printf (0040d760)
0040D891   add         esp,4
0040D894   jmp         $L549+2Bh (0040d91e)
0040D899   push        offset string "303" (00422fbc)
0040D89E   call        printf (0040d760)
0040D8A3   add         esp,4
0040D8A6   jmp         $L549+2Bh (0040d91e)
0040D8A8   push        offset string "304" (00422fb8)
0040D8AD   call        printf (0040d760)
0040D8B2   add         esp,4
0040D8B5   jmp         $L549+2Bh (0040d91e)
0040D8B7   push        offset string "305" (00422fb4)
0040D8BC   call        printf (0040d760)
0040D8C1   add         esp,4
0040D8C4   jmp         $L549+2Bh (0040d91e)
$L543:
0040D8C6   push        offset string "306" (00422fb0)
0040D8CB   call        printf (0040d760)
0040D8D0   add         esp,4
0040D8D3   jmp         $L549+2Bh (0040d91e)
$L545:
0040D8D5   push        offset string "307" (00422fac)
0040D8DA   call        printf (0040d760)
0040D8DF   add         esp,4
0040D8E2   jmp         $L549+2Bh (0040d91e)
$L547:
0040D8E4   push        offset string "308" (00422fa8)
0040D8E9   call        printf (0040d760)
0040D8EE   add         esp,4
0040D8F1   jmp         $L549+2Bh (0040d91e)
$L549:
0040D8F3   push        offset string "309" (00422fa4)
0040D8F8   call        printf (0040d760)
0040D8FD   add         esp,4
0040D900   jmp         $L549+2Bh (0040d91e)
0040D902   push        offset string "3" (0042210c)
0040D907   call        printf (0040d760)
0040D90C   add         esp,4
0040D90F   jmp         $L549+2Bh (0040d91e)
0040D911   push        offset string "None" (00422f6c)
0040D916   call        printf (0040d760)
0040D91B   add         esp,4
0040D91E   pop         edi
0040D91F   pop         esi
0040D920   pop         ebx
0040D921   add         esp,44h
0040D924   cmp         ebp,esp
0040D926   call        __chkesp (004010a0)
0040D92B   mov         esp,ebp
0040D92D   pop         ebp
0040D92E   ret

Memory：

0040D92F  C6 D8 40 00  曝@.
0040D933  D5 D8 40 00  肇@.
0040D937  E4 D8 40 00  湄@.
0040D93B  F3 D8 40 00  筘@.

小结：

此时的结构称为树型结构，间断大于256时的一种反汇编结构，类似于二分查找

总结

switch适用于其值为连续近似的情况，在此情况下switch的性能要高于if多分支，其因为switch利用索引的原理，完成以空间换时间。在不同的情况下，switch的反汇编会有所不同，大致分为四种情况，1.类似于if；2.使用大表结构；3.使用大表+小表结构；4.使用树型结构。

 
switch语句反汇编

正常switch语句在VC6++的编译器环境下，3个或者3个以下的条件，反汇编语句跟if else if ......没啥区别，也是一个一个比较，看下面例子

源代码

#include "stdafx.h"    
#include <string.h>

void fun(int x)
{
    switch (x)
    {
    case 1:
        printf("1\n");
        break;
    case 2:
        printf("2\n");
        break;
    case 3:
        printf("3\n");
        break;
    default:
        printf("Error\n");
        break;
    }
}

int main(int argc, char* argv[])
{
    fun(2);
    return 0;
}            

反汇编代码

4:    #include "stdafx.h"
5:    #include <string.h>
6:
7:    void fun(int x)
8:    {
00401020   push        ebp
00401021   mov         ebp,esp
00401023   sub         esp,44h
00401026   push        ebx
00401027   push        esi
00401028   push        edi
00401029   lea         edi,[ebp-44h]
0040102C   mov         ecx,11h
00401031   mov         eax,0CCCCCCCCh
00401036   rep stos    dword ptr [edi]
9:        switch (x)
10:       {
00401038   mov         eax,dword ptr [ebp+8]
0040103B   mov         dword ptr [ebp-4],eax
0040103E   cmp         dword ptr [ebp-4],1
00401042   je          fun+32h (00401052)
00401044   cmp         dword ptr [ebp-4],2
00401048   je          fun+41h (00401061)
0040104A   cmp         dword ptr [ebp-4],3
0040104E   je          fun+50h (00401070)
00401050   jmp         fun+5Fh (0040107f)
11:       case 1:
12:           printf("1\n");
00401052   push        offset string "1\n" (0042002c)
00401057   call        printf (00401110)
0040105C   add         esp,4
13:           break;
0040105F   jmp         fun+6Ch (0040108c)
14:       case 2:
15:           printf("2\n");
00401061   push        offset string "2\n" (00420028)
00401066   call        printf (00401110)
0040106B   add         esp,4
16:           break;
0040106E   jmp         fun+6Ch (0040108c)
17:       case 3:
18:           printf("3\n");
00401070   push        offset string "3\n" (00420024)
00401075   call        printf (00401110)
0040107A   add         esp,4
19:           break;
0040107D   jmp         fun+6Ch (0040108c)
20:       //case 4:
21:       //  printf("4\n");
22:       //  break;
23:       default:
24:           printf("Error\n");
0040107F   push        offset string "Error\n" (0042001c)
00401084   call        printf (00401110)
00401089   add         esp,4
25:           break;
26:       }
27:   }
0040108C   pop         edi
0040108D   pop         esi
0040108E   pop         ebx
0040108F   add         esp,44h
00401092   cmp         ebp,esp
00401094   call        __chkesp (00401190)
00401099   mov         esp,ebp
0040109B   pop         ebp
0040109C   ret


1.添加case后面的值，一个一个增加，观察反汇编代码的变化(何时生成大表).    

通过测试，在switch语句添加到含有4个case语句的时候就开始生成大表，且case语句后面的常量是连续的；

源代码

#include "stdafx.h"    
#include <string.h>

void fun(int x)
{
    switch (x)
    {
    case 1:
        printf("1\n");
        break;
    case 2:
        printf("2\n");
        break;
    case 3:
        printf("3\n");
        break;
    case 4:
        printf("4\n");
        break;
    default:
        printf("Error\n");
        break;
    }
}

int main(int argc, char* argv[])
{
    fun(2);
    return 0;
}

反汇编代码

4:    #include "stdafx.h"
5:    #include <string.h>
6:
7:    void fun(int x)
8:    {
0040B800   push        ebp
0040B801   mov         ebp,esp
0040B803   sub         esp,44h
0040B806   push        ebx
0040B807   push        esi
0040B808   push        edi
0040B809   lea         edi,[ebp-44h]
0040B80C   mov         ecx,11h
0040B811   mov         eax,0CCCCCCCCh
0040B816   rep stos    dword ptr [edi]
9:        switch (x)
10:       {
0040B818   mov         eax,dword ptr [ebp+8]
0040B81B   mov         dword ptr [ebp-4],eax
0040B81E   mov         ecx,dword ptr [ebp-4]
0040B821   sub         ecx,1
0040B824   mov         dword ptr [ebp-4],ecx
0040B827   cmp         dword ptr [ebp-4],3
0040B82B   ja          $L803+0Fh (0040b873)
0040B82D   mov         edx,dword ptr [ebp-4]
0040B830   jmp         dword ptr [edx*4+40B891h]
11:       case 1:
12:           printf("1\n");
0040B837   push        offset string "1\n" (0042013c)
0040B83C   call        printf (00401110)
0040B841   add         esp,4
13:           break;
0040B844   jmp         $L803+1Ch (0040b880)
14:       case 2:
15:           printf("2\n");
0040B846   push        offset string "2\n" (0042002c)
0040B84B   call        printf (00401110)
0040B850   add         esp,4
16:           break;
0040B853   jmp         $L803+1Ch (0040b880)
17:       case 3:
18:           printf("3\n");
0040B855   push        offset string "3\n" (00420028)
0040B85A   call        printf (00401110)
0040B85F   add         esp,4
19:           break;
0040B862   jmp         $L803+1Ch (0040b880)
20:       case 4:
21:           printf("4\n");
0040B864   push        offset string "4\n" (00420024)
0040B869   call        printf (00401110)
0040B86E   add         esp,4
22:           break;
0040B871   jmp         $L803+1Ch (0040b880)
23:       default:
24:           printf("Error\n");
0040B873   push        offset string "Error\n" (0042001c)
0040B878   call        printf (00401110)
0040B87D   add         esp,4
25:           break;
26:       }
27:   }
0040B880   pop         edi
0040B881   pop         esi
0040B882   pop         ebx
0040B883   add         esp,44h
0040B886   cmp         ebp,esp
0040B888   call        __chkesp (00401190)
0040B88D   mov         esp,ebp
0040B88F   pop         ebp
0040B890   ret
0040B891   aaa
0040B892   mov         eax,0B8460040h
0040B897   inc         eax
0040B898   add         byte ptr [ebp-48h],dl
0040B89B   inc         eax
0040B89C   add         byte ptr [eax+edi*4+40h],ah
0040B8A0   add         ah,cl

对应生成的大表

0040B891  37 B8 40 00
0040B895  46 B8 40 00
0040B899  55 B8 40 00
0040B89D  64 B8 40 00                

上述的大表刚好是4个，这也对应了是4个case语句；
                
对应计算公式的反汇编代码

0040B818   mov         eax,dword ptr [ebp+8]
0040B81B   mov         dword ptr [ebp-4],eax
0040B81E   mov         ecx,dword ptr [ebp-4]
0040B821   sub         ecx,1
0040B824   mov         dword ptr [ebp-4],ecx
0040B827   cmp         dword ptr [ebp-4],3
0040B82B   ja          $L803+0Fh (0040b873)
0040B82D   mov         edx,dword ptr [ebp-4]
0040B830   jmp         dword ptr [edx*4+40B891h]

[edx*4+40B891h]  1x4=4 4+40B891 ----> 刚好就是上面大表对应的0040B895地址，里面的内容就是直达返回结果的内存地址编号0040B846
                                    
                                    
2.将1中的常量值的顺序打乱，观察反汇编代码(观察顺序是否会影响生成大表).        

打乱case语句顺序的情景

源代码
#include "stdafx.h"    
#include <string.h>

void fun(int x)
{
    switch (x)
    {
    case 3:
        printf("1\n");
        break;
    case 5:
        printf("2\n");
        break;
    case 7:
        printf("3\n");
        break;
    case 4:
        printf("4\n");
        break;
    case 2:
        printf("5\n");
        break;
    case 1:
        printf("6\n");
        break;
    case 6:
        printf("7\n");
        break;
    case 8:
        printf("8\n");
        break;
    default:
        printf("Error\n");
        break;
    }
}

int main(int argc, char* argv[])
{
    fun(2);
    return 0;
}            


反汇编代码

4:    #include "stdafx.h"
5:    #include <string.h>
6:
7:    void fun(int x)
8:    {
0040B800   push        ebp
0040B801   mov         ebp,esp
0040B803   sub         esp,44h
0040B806   push        ebx
0040B807   push        esi
0040B808   push        edi
0040B809   lea         edi,[ebp-44h]
0040B80C   mov         ecx,11h
0040B811   mov         eax,0CCCCCCCCh
0040B816   rep stos    dword ptr [edi]
9:        switch (x)
10:       {
0040B818   mov         eax,dword ptr [ebp+8]
0040B81B   mov         dword ptr [ebp-4],eax
0040B81E   mov         ecx,dword ptr [ebp-4]
0040B821   sub         ecx,1
0040B824   mov         dword ptr [ebp-4],ecx
0040B827   cmp         dword ptr [ebp-4],7
0040B82B   ja          $L811+0Fh (0040b8b3)
0040B831   mov         edx,dword ptr [ebp-4]
0040B834   jmp         dword ptr [edx*4+40B8D1h]
11:       case 3:
12:           printf("1\n");
0040B83B   push        offset string "1\n" (00420f8c)
0040B840   call        printf (00401110)
0040B845   add         esp,4
13:           break;
0040B848   jmp         $L811+1Ch (0040b8c0)
14:       case 5:
15:           printf("2\n");
0040B84A   push        offset string "2\n" (00420f88)
0040B84F   call        printf (00401110)
0040B854   add         esp,4
16:           break;
0040B857   jmp         $L811+1Ch (0040b8c0)
17:       case 7:
18:           printf("3\n");
0040B859   push        offset string "3\n" (00420f84)
0040B85E   call        printf (00401110)
0040B863   add         esp,4
19:           break;
0040B866   jmp         $L811+1Ch (0040b8c0)
20:       case 4:
21:           printf("4\n");
0040B868   push        offset string "4\n" (00420f34)
0040B86D   call        printf (00401110)
0040B872   add         esp,4
22:           break;
0040B875   jmp         $L811+1Ch (0040b8c0)
23:       case 2:
24:           printf("5\n");
0040B877   push        offset string "5\n" (0042013c)
0040B87C   call        printf (00401110)
0040B881   add         esp,4
25:           break;
0040B884   jmp         $L811+1Ch (0040b8c0)
26:       case 1:
27:           printf("6\n");
0040B886   push        offset string "6\n" (0042002c)
0040B88B   call        printf (00401110)
0040B890   add         esp,4
28:           break;
0040B893   jmp         $L811+1Ch (0040b8c0)
29:       case 6:
30:           printf("7\n");
0040B895   push        offset string "7\n" (00420028)
0040B89A   call        printf (00401110)
0040B89F   add         esp,4
31:           break;
0040B8A2   jmp         $L811+1Ch (0040b8c0)
32:       case 8:
33:           printf("8\n");
0040B8A4   push        offset string "8\n" (00420024)
0040B8A9   call        printf (00401110)
0040B8AE   add         esp,4
34:           break;
0040B8B1   jmp         $L811+1Ch (0040b8c0)
35:       default:
36:           printf("Error\n");
0040B8B3   push        offset string "Error\n" (0042001c)
0040B8B8   call        printf (00401110)
0040B8BD   add         esp,4
37:           break;
38:       }
39:   }
0040B8C0   pop         edi
0040B8C1   pop         esi
0040B8C2   pop         ebx
0040B8C3   add         esp,44h
0040B8C6   cmp         ebp,esp
0040B8C8   call        __chkesp (00401190)
0040B8CD   mov         esp,ebp
0040B8CF   pop         ebp
0040B8D0   ret
0040B8D1   xchg        bh,byte ptr [eax-4788FFC0h]
0040B8D7   inc         eax
0040B8D8   add         byte ptr [ebx],bh
0040B8DA   mov         eax,0B8680040h
0040B8DF   inc         eax
0040B8E0   add         byte ptr [edx-48h],cl
0040B8E3   inc         eax
0040B8E4   add         byte ptr [ebp+590040B8h],dl
0040B8EA   mov         eax,0B8A40040h
0040B8EF   inc         eax
0040B8F0   add         ah,cl


对应生成的大表

0040B8D1  86 B8 40 00  
0040B8D5  77 B8 40 00
0040B8D9  3B B8 40 00
0040B8DD  68 B8 40 00
0040B8E1  4A B8 40 00
0040B8E5  95 B8 40 00
0040B8E9  59 B8 40 00
0040B8ED  A4 B8 40 00                    
                                    
                                    
得出结论就是，即使case语句的常量打乱了顺序，仍然不影响大表的生成

                                    
3.将case后面的值改成从100开始到109，观察汇编变化(观察值较大时是否生成大表).                    

源代码

#include "stdafx.h"    
#include <string.h>

void fun(int x)
{
    switch (x)
    {
    case 100:
        printf("100\n");
        break;
    case 101:
        printf("101\n");
        break;
    case 102:
        printf("102\n");
        break;
    case 103:
        printf("103\n");
        break;
    case 104:
        printf("104\n");
        break;
    case 105:
        printf("105\n");
        break;
    case 106:
        printf("106\n");
        break;
    case 107:
        printf("107\n");
        break;
    case 108:
        printf("108\n");
        break;
    case 109:
        printf("109\n");
        break;
    default:
        printf("Error\n");
        break;
    }
}

int main(int argc, char* argv[])
{
    fun(106);
    return 0;
}


反汇编代码

4:    #include "stdafx.h"
5:    #include <string.h>
6:
7:    void fun(int x)
8:    {
0040B800   push        ebp
0040B801   mov         ebp,esp
0040B803   sub         esp,44h
0040B806   push        ebx
0040B807   push        esi
0040B808   push        edi
0040B809   lea         edi,[ebp-44h]
0040B80C   mov         ecx,11h
0040B811   mov         eax,0CCCCCCCCh
0040B816   rep stos    dword ptr [edi]
9:        switch (x)
10:       {
0040B818   mov         eax,dword ptr [ebp+8]
0040B81B   mov         dword ptr [ebp-4],eax
0040B81E   mov         ecx,dword ptr [ebp-4]
0040B821   sub         ecx,64h
0040B824   mov         dword ptr [ebp-4],ecx
0040B827   cmp         dword ptr [ebp-4],9
0040B82B   ja          $L815+0Fh (0040b8d7)
0040B831   mov         edx,dword ptr [ebp-4]
0040B834   jmp         dword ptr [edx*4+40B8F5h]
11:       case 100:
12:           printf("100\n");
0040B83B   push        offset string "100\n" (00420fc8)
0040B840   call        printf (00401110)
0040B845   add         esp,4
13:           break;
0040B848   jmp         $L815+1Ch (0040b8e4)
14:       case 101:
15:           printf("101\n");
0040B84D   push        offset string "101\n" (00420fc0)
0040B852   call        printf (00401110)
0040B857   add         esp,4
16:           break;
0040B85A   jmp         $L815+1Ch (0040b8e4)
17:       case 102:
18:           printf("102\n");
0040B85F   push        offset string "102\n" (00420fb8)
0040B864   call        printf (00401110)
0040B869   add         esp,4
19:           break;
0040B86C   jmp         $L815+1Ch (0040b8e4)
20:       case 103:
21:           printf("103\n");
0040B86E   push        offset string "103\n" (00420f84)
0040B873   call        printf (00401110)
0040B878   add         esp,4
22:           break;
0040B87B   jmp         $L815+1Ch (0040b8e4)
23:       case 104:
24:           printf("104\n");
0040B87D   push        offset string "104\n" (00420fb0)
0040B882   call        printf (00401110)
0040B887   add         esp,4
25:           break;
0040B88A   jmp         $L815+1Ch (0040b8e4)
26:       case 105:
27:           printf("105\n");
0040B88C   push        offset string "105\n" (00420fa8)
0040B891   call        printf (00401110)
0040B896   add         esp,4
28:           break;
0040B899   jmp         $L815+1Ch (0040b8e4)
29:       case 106:
30:           printf("106\n");
0040B89B   push        offset string "106\n" (00420fa0)
0040B8A0   call        printf (00401110)
0040B8A5   add         esp,4
31:           break;
0040B8A8   jmp         $L815+1Ch (0040b8e4)
32:       case 107:
33:           printf("107\n");
0040B8AA   push        offset string "107\n" (00420f98)
0040B8AF   call        printf (00401110)
0040B8B4   add         esp,4
34:           break;
0040B8B7   jmp         $L815+1Ch (0040b8e4)
35:       case 108:
36:           printf("108\n");
0040B8B9   push        offset string "108\n" (00420024)
0040B8BE   call        printf (00401110)
0040B8C3   add         esp,4
37:           break;
0040B8C6   jmp         $L815+1Ch (0040b8e4)
38:       case 109:
39:           printf("109\n");
0040B8C8   push        offset string "109\n" (00420f90)
0040B8CD   call        printf (00401110)
0040B8D2   add         esp,4
40:           break;
0040B8D5   jmp         $L815+1Ch (0040b8e4)
41:       default:
42:           printf("Error\n");
0040B8D7   push        offset string "Error\n" (0042001c)
0040B8DC   call        printf (00401110)
0040B8E1   add         esp,4
43:           break;
44:       }
45:   }
0040B8E4   pop         edi
0040B8E5   pop         esi
0040B8E6   pop         ebx
0040B8E7   add         esp,44h
0040B8EA   cmp         ebp,esp
0040B8EC   call        __chkesp (00401190)
0040B8F1   mov         esp,ebp
0040B8F3   pop         ebp
0040B8F4   ret
0040B8F5   cmp         edi,dword ptr [eax-47B2FFC0h]
0040B8FB   inc         eax
0040B8FC   add         byte ptr [edi-48h],bl
0040B8FF   inc         eax
0040B900   add         byte ptr [esi-48h],ch
0040B903   inc         eax
0040B904   add         byte ptr [ebp-48h],bh
0040B907   inc         eax
0040B908   add         byte ptr [eax+edi*4-4764FFC0h],cl
0040B90F   inc         eax
0040B910   add         byte ptr [edx-46FFBF48h],ch
0040B916   mov         eax,0B8C80040h
0040B91B   inc         eax
0040B91C   add         ah,cl

根据大表计算

64 -- 十进制 100
106-100 = 6
edx*4+40B8F5h
从上面计算公式得出
6×4+40B8F5 意思就是0040B8F5所在的大表向下以4个字节移动6位

大表
0040B8F5  3B B8 40 00
0040B8F9  4D B8 40 00
0040B8FD  5F B8 40 00
0040B901  6E B8 40 00
0040B905  7D B8 40 00
0040B909  8C B8 40 00
0040B90D  9B B8 40 00
0040B911  AA B8 40 00
0040B915  B9 B8 40 00
0040B919  C8 B8 40 00
                                    
向下以4字节移动6位刚好就是大表地址 0040B90D  9B B8 40 00  对应的内存地址: 0040B89B

30:           printf("106\n");
0040B89B   push        offset string "106\n" (00420fa0)
0040B8A0   call        printf (00401110)
0040B8A5   add         esp,4

刚好打印106

所以此处结论是依然会生成大表

                                    
4.将连续的10项中抹去1项或者2项，观察反汇编有无变化(观察大表空缺位置的处理).                

抹去其中case语句 108

#include "stdafx.h"    
#include <string.h>

void fun(int x)
{
    switch (x)
    {
    case 100:
        printf("100\n");
        break;
    case 101:
        printf("101\n");
        break;
    case 102:
        printf("102\n");
        break;
    case 103:
        printf("103\n");
        break;
    case 104:
        printf("104\n");
        break;
    case 105:
        printf("105\n");
        break;
    case 106:
        printf("106\n");
        break;
    case 107:
        printf("107\n");
        break;

    case 109:
        printf("109\n");
        break;
    default:
        printf("Error\n");
        break;
    }
}

int main(int argc, char* argv[])
{
    fun(106);
    return 0;
}            


反汇编代码

4:    #include "stdafx.h"
5:    #include <string.h>
6:
7:    void fun(int x)
8:    {
0040B800   push        ebp
0040B801   mov         ebp,esp
0040B803   sub         esp,44h
0040B806   push        ebx
0040B807   push        esi
0040B808   push        edi
0040B809   lea         edi,[ebp-44h]
0040B80C   mov         ecx,11h
0040B811   mov         eax,0CCCCCCCCh
0040B816   rep stos    dword ptr [edi]
9:        switch (x)
10:       {
0040B818   mov         eax,dword ptr [ebp+8]
0040B81B   mov         dword ptr [ebp-4],eax
0040B81E   mov         ecx,dword ptr [ebp-4]
0040B821   sub         ecx,64h
0040B824   mov         dword ptr [ebp-4],ecx
0040B827   cmp         dword ptr [ebp-4],9
0040B82B   ja          $L813+0Fh (0040b8c5)
0040B831   mov         edx,dword ptr [ebp-4]
0040B834   jmp         dword ptr [edx*4+40B8E3h]
11:       case 100:
12:           printf("100\n");
0040B83B   push        offset string "100\n" (00420fc0)
0040B840   call        printf (00401110)
0040B845   add         esp,4
13:           break;
0040B848   jmp         $L813+1Ch (0040b8d2)
14:       case 101:
15:           printf("101\n");
0040B84D   push        offset string "101\n" (00420fb8)
0040B852   call        printf (00401110)
0040B857   add         esp,4
16:           break;
0040B85A   jmp         $L813+1Ch (0040b8d2)
17:       case 102:
18:           printf("102\n");
0040B85C   push        offset string "3\n" (00420f84)
0040B861   call        printf (00401110)
0040B866   add         esp,4
19:           break;
0040B869   jmp         $L813+1Ch (0040b8d2)
20:       case 103:
21:           printf("103\n");
0040B86B   push        offset string "103\n" (00420fb0)
0040B870   call        printf (00401110)
0040B875   add         esp,4
22:           break;
0040B878   jmp         $L813+1Ch (0040b8d2)
23:       case 104:
24:           printf("104\n");
0040B87A   push        offset string "104\n" (00420fa8)
0040B87F   call        printf (00401110)
0040B884   add         esp,4
25:           break;
0040B887   jmp         $L813+1Ch (0040b8d2)
26:       case 105:
27:           printf("105\n");
0040B889   push        offset string "105\n" (00420fa0)
0040B88E   call        printf (00401110)
0040B893   add         esp,4
28:           break;
0040B896   jmp         $L813+1Ch (0040b8d2)
29:       case 106:
30:           printf("106\n");
0040B898   push        offset string "106\n" (00420f98)
0040B89D   call        printf (00401110)
0040B8A2   add         esp,4
31:           break;
0040B8A5   jmp         $L813+1Ch (0040b8d2)
32:       case 107:
33:           printf("107\n");
0040B8A7   push        offset string "8\n" (00420024)
0040B8AC   call        printf (00401110)
0040B8B1   add         esp,4
34:           break;
0040B8B4   jmp         $L813+1Ch (0040b8d2)
35:
36:       case 109:
37:           printf("109\n");
0040B8B6   push        offset string "109\n" (00420f90)
0040B8BB   call        printf (00401110)
0040B8C0   add         esp,4
38:           break;
0040B8C3   jmp         $L813+1Ch (0040b8d2)
39:       default:
40:           printf("Error\n");
0040B8C5   push        offset string "Error\n" (0042001c)
0040B8CA   call        printf (00401110)
0040B8CF   add         esp,4
41:           break;
42:       }
43:   }
0040B8D2   pop         edi
0040B8D3   pop         esi
0040B8D4   pop         ebx
0040B8D5   add         esp,44h
0040B8D8   cmp         ebp,esp
0040B8DA   call        __chkesp (00401190)
0040B8DF   mov         esp,ebp
0040B8E1   pop         ebp
0040B8E2   ret
0040B8E3   cmp         edi,dword ptr [eax-47B2FFC0h]
0040B8E9   inc         eax
0040B8EA   add         byte ptr [eax+edi*4+40h],bl
0040B8EE   add         byte ptr [ebx-48h],ch
0040B8F1   inc         eax
0040B8F2   add         byte ptr [edx-48h],bh
0040B8F5   inc         eax
0040B8F6   add         byte ptr [ecx-67FFBF48h],cl
0040B8FC   mov         eax,0B8A70040h
0040B901   inc         eax
0040B902   add         ch,al
0040B904   mov         eax,0B8B60040h
0040B909   inc         eax
0040B90A   add         ah,cl


[edx*4+40B8E3h] 计算公式操作跟上面一样，其中有一个区别就是去掉了一个case语句会将default的对应内存地址放入这个大表中，下面所在位置
0040B903  C5 B8 40 00

大表

0040B8E3  3B B8 40 00
0040B8E7  4D B8 40 00
0040B8EB  5C B8 40 00
0040B8EF  6B B8 40 00
0040B8F3  7A B8 40 00
0040B8F7  89 B8 40 00
0040B8FB  98 B8 40 00
0040B8FF  A7 B8 40 00
0040B903  C5 B8 40 00
0040B907  B6 B8 40 00                     
                                    
得出结论：
(1)当抹去1个case语句会将default的内存地址编号放入大表中；
(2)当抹去2个case语句会将default的内存地址编号放入大表中,此时就有两个相同的default内存地址在大表中；
(3)当抹去3个case语句会将default的内存地址编号放入大表中,此时就有三个相同的default内存地址在大表中；
(4)当抹去4个case语句会将default的内存地址编号放入大表中,此时就有四个相同的default内存地址在大表中；
(5)当抹去5个case语句会将default的内存地址编号放入大表中,此时就有五个相同的default内存地址在大表中；
(6)当抹去6个case语句的时候会生成小表；
                                    
5.在10项中连续抹去，不要抹去最大值和最小值(观察何时生成小表).        

上述4例题中已经论证，当抹去6个case语句的时候会生成小表；    

源代码


反汇编代码

4:    #include "stdafx.h"
5:    #include <string.h>
6:
7:    void fun(int x)
8:    {
0040B800   push        ebp
0040B801   mov         ebp,esp
0040B803   sub         esp,44h
0040B806   push        ebx
0040B807   push        esi
0040B808   push        edi
0040B809   lea         edi,[ebp-44h]
0040B80C   mov         ecx,11h
0040B811   mov         eax,0CCCCCCCCh
0040B816   rep stos    dword ptr [edi]
9:        switch (x)
10:       {
0040B818   mov         eax,dword ptr [ebp+8]
0040B81B   mov         dword ptr [ebp-4],eax
0040B81E   mov         ecx,dword ptr [ebp-4]
0040B821   sub         ecx,64h
0040B824   mov         dword ptr [ebp-4],ecx
0040B827   cmp         dword ptr [ebp-4],9
0040B82B   ja          $L803+0Fh (0040b87b)
0040B82D   mov         eax,dword ptr [ebp-4]
0040B830   xor         edx,edx
0040B832   mov         dl,byte ptr  (0040b8ad)[eax]
0040B838   jmp         dword ptr [edx*4+40B899h]
11:       case 100:
12:           printf("100\n");
0040B83F   push        offset string "100\n" (00420fa0)
0040B844   call        printf (00401110)
0040B849   add         esp,4
13:           break;
0040B84C   jmp         $L803+1Ch (0040b888)
14:       case 101:
15:           printf("101\n");
0040B84E   push        offset string "101\n" (00420f98)
0040B853   call        printf (00401110)
0040B858   add         esp,4
16:           break;
0040B85B   jmp         $L803+1Ch (0040b888)
17:       case 102:
18:           printf("102\n");
0040B85D   push        offset string "104\n" (00420024)
0040B862   call        printf (00401110)
0040B867   add         esp,4
19:           break;
0040B86A   jmp         $L803+1Ch (0040b888)
20:
21:
22:       case 109:
23:           printf("109\n");
0040B86C   push        offset string "109\n" (00420f90)
0040B871   call        printf (00401110)
0040B876   add         esp,4
24:           break;
0040B879   jmp         $L803+1Ch (0040b888)
25:       default:
26:           printf("Error\n");
0040B87B   push        offset string "Error\n" (0042001c)
0040B880   call        printf (00401110)
0040B885   add         esp,4
27:           break;
28:       }
29:   }
0040B888   pop         edi
0040B889   pop         esi
0040B88A   pop         ebx
0040B88B   add         esp,44h
0040B88E   cmp         ebp,esp
0040B890   call        __chkesp (00401190)
0040B895   mov         esp,ebp
0040B897   pop         ebp
0040B898   ret
0040B899   aas
0040B89A   mov         eax,0B84E0040h
0040B89F   inc         eax
0040B8A0   add         byte ptr [ebp-48h],bl
0040B8A3   inc         eax
0040B8A4   add         byte ptr [eax+edi*4+40h],ch
0040B8A8   add         byte ptr [ebx-48h],bh
0040B8AB   inc         eax
0040B8AC   add         byte ptr [eax],al
0040B8AE   add         dword ptr [edx],eax
0040B8B0   add         al,4
0040B8B2   add         al,4
0040B8B4   add         al,4
0040B8B6   add         ecx,esp

对应的大表和小表

0040B899  3F B8 40 00
0040B89D  4E B8 40 00
0040B8A1  5D B8 40 00
0040B8A5  6C B8 40 00
0040B8A9  7B B8 40 00
0040B8AD  00 01 02 04   //小表
0040B8B1  04 04 04 04   //小表
0040B8B5  04 03 CC CC      //小表

其中反汇编 0040B832   mov         dl,byte ptr  (0040b8ad)[eax] 含义

将小表0040B8AD加上eax的值，就是结果，而eax值此时是6，所以向后移动6位即可

0040B8AD  00 01 02 04
0040B8B1  04 04 04 04   

00 01 02 04 04 04 04 04 ---> 向后移动6位，就是数6位，得到的结果是04，所以edx的结果是4 根据下面的公式

edx*4+40B899h  4×4+0040B899  可以得出结果是以4字节向后4位即可得到计算结果的内存地址编号：0040B8A9  7B B8 40 00

内存编号对应的内存地址：0040B87B

26:           printf("Error\n");
0040B87B   push        offset string "Error\n" (0042001c)
0040B880   call        printf (00401110)
0040B885   add         esp,4
                                    
6.将case后面常量表达式改成毫不连续的值，观察反汇编变化.                                    

源代码
#include "stdafx.h"    
#include <string.h>

void fun(int x)
{
    switch (x)
    {
    case 10:
        printf("100\n");
        break;
    case 201:
        printf("101\n");
        break;
    case 2:
        printf("102\n");
        break;
    case 692:
        printf("102\n");
        break;
    case 532:
        printf("102\n");
        break;
    case 308:
        printf("102\n");
        break;
    case 804:
        printf("102\n");
        break;
    case 749:
        printf("102\n");
        break;
    case 1033:
        printf("102\n");
        break;

    case 29:
        printf("109\n");
        break;
    default:
        printf("Error\n");
        break;
    }
}

int main(int argc, char* argv[])
{
    fun(308);
    return 0;
}            


反汇编代码

4:    #include "stdafx.h"
5:    #include <string.h>
6:
7:    void fun(int x)
8:    {
0040B800   push        ebp
0040B801   mov         ebp,esp
0040B803   sub         esp,44h
0040B806   push        ebx
0040B807   push        esi
0040B808   push        edi
0040B809   lea         edi,[ebp-44h]
0040B80C   mov         ecx,11h
0040B811   mov         eax,0CCCCCCCCh
0040B816   rep stos    dword ptr [edi]
9:        switch (x)
10:       {
0040B818   mov         eax,dword ptr [ebp+8]
0040B81B   mov         dword ptr [ebp-4],eax
0040B81E   cmp         dword ptr [ebp-4],214h
0040B825   jg          fun+74h (0040b874)
0040B827   cmp         dword ptr [ebp-4],214h
0040B82E   je          fun+0F9h (0040b8f9)
0040B834   cmp         dword ptr [ebp-4],1Dh
0040B838   jg          fun+59h (0040b859)
0040B83A   cmp         dword ptr [ebp-4],1Dh
0040B83E   je          fun+144h (0040b944)
0040B844   cmp         dword ptr [ebp-4],2
0040B848   je          fun+0DBh (0040b8db)
0040B84E   cmp         dword ptr [ebp-4],0Ah
0040B852   je          fun+0B7h (0040b8b7)
0040B854   jmp         fun+153h (0040b953)
0040B859   cmp         dword ptr [ebp-4],0C9h
0040B860   je          fun+0C9h (0040b8c9)
0040B862   cmp         dword ptr [ebp-4],134h
0040B869   je          fun+108h (0040b908)
0040B86F   jmp         fun+153h (0040b953)
0040B874   cmp         dword ptr [ebp-4],324h
0040B87B   jg          fun+0A5h (0040b8a5)
0040B87D   cmp         dword ptr [ebp-4],324h
0040B884   je          fun+117h (0040b917)
0040B88A   cmp         dword ptr [ebp-4],2B4h
0040B891   je          fun+0EAh (0040b8ea)
0040B893   cmp         dword ptr [ebp-4],2EDh
0040B89A   je          fun+126h (0040b926)
0040B8A0   jmp         fun+153h (0040b953)
0040B8A5   cmp         dword ptr [ebp-4],409h
0040B8AC   je          fun+135h (0040b935)
0040B8B2   jmp         fun+153h (0040b953)
11:       case 10:
12:           printf("100\n");
0040B8B7   push        offset string "100\n" (00420fa0)
0040B8BC   call        printf (00401110)
0040B8C1   add         esp,4
13:           break;
0040B8C4   jmp         fun+160h (0040b960)
14:       case 201:
15:           printf("101\n");
0040B8C9   push        offset string "101\n" (00420f98)
0040B8CE   call        printf (00401110)
0040B8D3   add         esp,4
16:           break;
0040B8D6   jmp         fun+160h (0040b960)
17:       case 2:
18:           printf("102\n");
0040B8DB   push        offset string "104\n" (00420024)
0040B8E0   call        printf (00401110)
0040B8E5   add         esp,4
19:           break;
0040B8E8   jmp         fun+160h (0040b960)
20:       case 692:
21:           printf("102\n");
0040B8EA   push        offset string "104\n" (00420024)
0040B8EF   call        printf (00401110)
0040B8F4   add         esp,4
22:           break;
0040B8F7   jmp         fun+160h (0040b960)
23:       case 532:
24:           printf("102\n");
0040B8F9   push        offset string "104\n" (00420024)
0040B8FE   call        printf (00401110)
0040B903   add         esp,4
25:           break;
0040B906   jmp         fun+160h (0040b960)
26:       case 308:
27:           printf("102\n");
0040B908   push        offset string "104\n" (00420024)
0040B90D   call        printf (00401110)
0040B912   add         esp,4
28:           break;
0040B915   jmp         fun+160h (0040b960)
29:       case 804:
30:           printf("102\n");
0040B917   push        offset string "104\n" (00420024)
0040B91C   call        printf (00401110)
0040B921   add         esp,4
31:           break;
0040B924   jmp         fun+160h (0040b960)
32:       case 749:
33:           printf("102\n");
0040B926   push        offset string "104\n" (00420024)
0040B92B   call        printf (00401110)
0040B930   add         esp,4
34:           break;
0040B933   jmp         fun+160h (0040b960)
35:       case 1033:
36:           printf("102\n");
0040B935   push        offset string "104\n" (00420024)
0040B93A   call        printf (00401110)
0040B93F   add         esp,4
37:           break;
0040B942   jmp         fun+160h (0040b960)
38:
39:       case 29:
40:           printf("109\n");
0040B944   push        offset string "109\n" (00420f90)
0040B949   call        printf (00401110)
0040B94E   add         esp,4
41:           break;
0040B951   jmp         fun+160h (0040b960)
42:       default:
43:           printf("Error\n");
0040B953   push        offset string "Error\n" (0042001c)
0040B958   call        printf (00401110)
0040B95D   add         esp,4
44:           break;
45:       }
46:   }
0040B960   pop         edi
0040B961   pop         esi
0040B962   pop         ebx
0040B963   add         esp,44h
0040B966   cmp         ebp,esp
0040B968   call        __chkesp (00401190)
0040B96D   mov         esp,ebp
0040B96F   pop         ebp
0040B970   ret


结论：此时switch语句不会生成大表和小表，跟if else if语句判断逻辑基本相同，不能提高效率，如果这样操作就没啥意义；

7.如果switch语句中的case语句常量是无序且每个数字之间大于256，那么这种情况也是无意义的，不能提高效率，这一种极端情况；

当分支数量大于3个且部分不连续的时候（差值较大）

switch ( argc )
{
case 10:
printf("case 10 ! \r\n");
break;
case 11:
printf("case 11 ! \r\n");
break;
case 12:
printf("case 12 ! \r\n");
break;
case 19:
printf("case 19 ! \r\n");
break;
default:
printf("default ! \r\n");
break;
}
getchar();

反汇编：

这个时候我们的两个值的差值是7。这个时候发现又不一样了，寻址方式变成了：movzx eax, byte ptr ds:[ eax + 0xE610A8 ]。dword 变 byte 了，我们数据窗体中看一下。如图选中内容，这就是小表。每一个元素仅仅占1个字节。小表大小也是最大值 19 - 最小值 10。接下来就是 jmp dword ptr ds:[ eax*4
+ 0xE61094 ]，这个当然，又是我们亲爱的大表了，联系上下文我们发现。小表里面存的就是大表的下标。为什么要这样设计呢？由于大表占四个字节，当差距比較大时，生成的大表自然也会变得非常大，这个时候使用小表，能够更加节约内存。

posted @ 2023-04-02 23:02 bonelee 阅读(545) 评论(0) 编辑收藏举报

刷新页面返回顶部

将者，智、信、仁、勇、严也。

Hi，我是李智华，华为-安全AI算法专家，欢迎来到安全攻防对抗的有趣世界。

C语言逆向分析——Switch语句，为何大多数情况较if语句更高效？就是因为查找表

Switch语句

Switch语句为什么高效

实例

case分支少的switch语句

case分支多的switch语句

case后的值打乱顺序

起始值对反汇编的影响

连续值中抹去少项

连续值中抹去多项

断开一定程度的选择

差值大的情况

总结

公告