x86---32汇编(2)---局部变量
在C/C++编程中,我们经常会用到局部变量,笔者想知道在汇编语言中是
如何使用局部变量的,根据《X86汇编语言》中的例子,才弄懂了汇编是如何
分配局部变量。用栈指针减去一个值(需要分配的内存大小)。
代码:
#include <stdio.h> #include <tchar.h> extern "C" void CalculateSums_(int a, int b, int c, int *s1, int *s2, int *s3); int _tmain(int argc, _TCHAR* argv[]) { int a = 3, b = 5, c = 8; int s1a, s2a, s3a; CalculateSums_(a, b, c, &s1a, &s2a, &s3a); //compute the sums again so we can verify the results //of CalculateSums_() int s1b = a + b + c; int s2b = a * a + b * b + c * c; int s3b = a * a*a + b * b*b + c * c*c; printf("Input:a:%4d b:%4d c:%4d\n", a, b, c); printf("Output:s1a:%4d s2a:%4d s3a:%4d\n", s1a, s2a, s3a); printf("s1b:%4d s2b:%4d s3b:%4d\n",s1b,s2b,s3b); return 0; }
.model flat,c .code ; extern "C" void CalculateSums_(int a, int b, int c, int* s1, int* s2, int* s3); ; ; Description: This function demonstrates a complete assembly ; language prolog and epilog. ; ; Returns: None. ; ; Computes: *s1 = a + b + c ; *s2 = a * a + b * b + c * c ; *s3 = a * a * a + b * b * b + c * c * c CalculateSums_ proc ;Function prolog push ebp mov ebp,esp sub esp,12 ;Allocate local storage space push ebx push esi push edi ;Load arguments mov eax,[ebp+8] ;eax="a" mov ebx,[ebp+12] ;ebx='b' mov ecx,[ebp+16] ;ecx='c' mov edx,[ebp+20] ;edx='s1' mov esi,[ebp+24] ;esi='s2' mov edi,[ebp+28] ;edi='s3' ;Compute 's1' mov [ebp-12],eax add [ebp-12],ebx add [ebp-12],ecx ;final 's1' result ;Compute 's2' imul eax,eax imul ebx,ebx imul ecx,ecx mov [ebp-8],eax add [ebp-8],ebx add [ebp-8],ecx ;final 's2' result ;Compute 's3' imul eax,[ebp+8] imul ebx,[ebp+12] imul ecx,[ebp+16] mov [ebp-4],eax add [ebp-4],ebx add [ebp-4],ecx ;final 's3' result ;Save 's1','s2','s3' mov eax,[ebp-12] mov [edx],eax ;save 's1' mov eax,[ebp-8] mov [esi],eax ;save 's2' mov eax,[ebp-4] mov [edi],eax ;save 's3' ;Function epilog pop edi pop esi pop ebx mov esp,ebp pop ebp ret CalculateSums_ endp end
在vs2017上的运行效果: