深入理解计算机系统_3e 第二章家庭作业 CS:APP3e chapter 2 homework

初始完成日期：2017.9.26

许可：除2.55对应代码外（如需使用请联系 randy.bryant@cs.cmu.edu），任何人可以自由的使用，修改，分发本文档的代码。

本机环境： （有一些需要在多种机器上测试的就没有试验）

frank@under:~/tmp$ uname -a
Linux under 4.10.0-35-generic #39~16.04.1-Ubuntu SMP Wed Sep 13 09:02:42 UTC 2017 x86_64 x86_64 x86_64 GNU/Linux

2.55

/*这一段代码的大部分来自http://csapp.cs.cmu.edu/3e/students.html*/
/* $begin show-bytes */
#include <stdio.h>
/* $end show-bytes */
#include <stdlib.h>
#include <string.h>
/* $begin show-bytes */

typedef unsigned char *byte_pointer;

void show_bytes(byte_pointer start, size_t len) {
    size_t i;
    for (i = 0; i < len; i++)
	printf(" %.2x", start[i]);    //line:data:show_bytes_printf
    printf("\n");
}

void show_int(int x) {
    show_bytes((byte_pointer) &x, sizeof(int)); //line:data:show_bytes_amp1
}

void show_float(float x) {
    show_bytes((byte_pointer) &x, sizeof(float)); //line:data:show_bytes_amp2
}

void show_pointer(void *x) {
    show_bytes((byte_pointer) &x, sizeof(void *)); //line:data:show_bytes_amp3
}
/* $end show-bytes */


/* $begin test-show-bytes */
void test_show_bytes(int val) {
    int ival = val;
    float fval = (float) ival;
    int *pval = &ival;
    show_int(ival);
    show_float(fval);
    show_pointer(pval);
}
/* $end test-show-bytes */

void simple_show_a() {
/* $begin simple-show-a */
int val = 0x87654321;
byte_pointer valp = (byte_pointer) &val;
show_bytes(valp, 1); /* A. */
show_bytes(valp, 2); /* B. */
show_bytes(valp, 3); /* C. */
/* $end simple-show-a */
}

void simple_show_b() {
/* $begin simple-show-b */
int val = 0x12345678;
byte_pointer valp = (byte_pointer) &val;
show_bytes(valp, 1); /* A. */
show_bytes(valp, 2); /* B. */
show_bytes(valp, 3); /* C. */
/* $end simple-show-b */
}

void float_eg() {
  int x = 3490593;
  float f = (float) x;
  printf("For x = %d\n", x);
  show_int(x);
  show_float(f);

  x = 3510593;
  f = (float) x;
  printf("For x = %d\n", x);
  show_int(x);
  show_float(f);

}

void string_ueg() {
/* $begin show-ustring */
const char *s = "ABCDEF";
show_bytes((byte_pointer) s, strlen(s)); 
/* $end show-ustring */
}

void string_leg() {
/* $begin show-lstring */
const char *s = "abcdef";
show_bytes((byte_pointer) s, strlen(s)); 
/* $end show-lstring */
}

void show_twocomp() 
{
/* $begin show-twocomp */
    short x = 12345; 
    short mx = -x; 
    
    show_bytes((byte_pointer) &x, sizeof(short)); 
    show_bytes((byte_pointer) &mx, sizeof(short)); 
/* $end show-twocomp */
}

int main(int argc, char *argv[])
{
    int val = 12345;

    if (argc > 1) {
	if (argc > 1) {
	    val = strtol(argv[1], NULL, 0);
	}
	printf("calling test_show_bytes\n");
	test_show_bytes(val);
    } else {
	printf("calling show_twocomp\n");
	show_twocomp();
	printf("Calling simple_show_a\n");
	simple_show_a();
	printf("Calling simple_show_b\n");
	simple_show_b();
	printf("Calling float_eg\n");
	float_eg();
	printf("Calling string_ueg\n");
	string_ueg();
	printf("Calling string_leg\n");
	string_leg();
    }
    return 0;
}

编译运行输出：

frank@under:~/tmp$ gcc 255.c && ./a.out 
calling show_twocomp
 39 30
 c7 cf
Calling simple_show_a
 21
 21 43
 21 43 65
Calling simple_show_b
 78
 78 56
 78 56 34
Calling float_eg
For x = 3490593
 21 43 35 00
 84 0c 55 4a
For x = 3510593
 41 91 35 00
 04 45 56 4a
Calling string_ueg
 41 42 43 44 45 46
Calling string_leg
 61 62 63 64 65 66

数据的低位放置在低地址处，字符串按照顺序从低位地址排列。由此可知该机器为小端字节排序。

2.56

show-bytes代码同2.55。

frank@under:~/tmp$ gcc 255.c && ./a.out 192837465
calling test_show_bytes
 59 77 7e 0b
 76 e7 37 4d
 28 00 4d 93 fc 7f 00 00

十进制192837465二进制表示为：1011011111100111011101011001。

十六进制表示为：0xB7E7759，所以第一行的数据表示这是小端字节排序。
将二进制小数点左移二十七位，由于单精度浮点数尾数部分只有23位（IEEE 756），多出来的4位1001将丢弃，由于默认的“round to even”，高位将进一位，即1.01101111110011101110110*2^27，bias = 2^7 - 1 = 127，所以阶码部分应为127+27=0x9A，整体为：0,10011010,01101111110011101110110即0x4D37E776。由此看出为小端字节排序。
该行为小端字节显示的指针。

2.57

#include <stdio.h>

typedef unsigned char *byte_pointer;

void show_short(void);
void show_long(void);
void show_double(void);
void show_bytes(byte_pointer start, size_t len);

int main(int argc, char const *argv[])
{
	show_short();
	show_long();
	show_double();
	return 0;
}

void show_bytes(byte_pointer start, size_t len) {
    size_t i;
    for (i = 0; i < len; i++)
	printf(" %.2x", start[i]);
    printf("\n");
}

void show_short(void)
{
	short i = 12345;
	printf("short i = 12345\n");
	show_bytes((byte_pointer)&i, sizeof i);
}
void show_long(void)
{
	long i = 123456789;
	printf("long i = 123456789\n");
	show_bytes((byte_pointer)&i, sizeof i);
}
void show_double(void)
{
	double i = 123456789.0;
	printf("double i = 123456789.0\n");
	show_bytes((byte_pointer)&i, sizeof i);
}

编译运行输出：

frank@under:~/tmp$ ./a.out 
short i = 12345
 39 30
long i = 123456789
 15 cd 5b 07 00 00 00 00
double i = 123456789.0
 00 00 00 54 34 6f 9d 41

2.58

#include <stdio.h>
#include <stdint.h>

int is_little_endian(void);

int main(int argc, char const *argv[])
{
	return is_little_endian();
}

int is_little_endian(void)
{
    int32_t i = 1;
	unsigned char *p = (unsigned char *)&i;
	if(*p)
	{
		return 1;
	}
	return 0;
}

编译运行输出：

frank@under:~/tmp$ gcc 258.c && ./a.out ; echo $?
1

2.59

#include <stdio.h>

int main(int argc, char const *argv[])
{
	int x = 0x89ABCDEF;
	int y = 0x76543210;
	printf("0x%.8X\n", x&0xFF | y&~0xFF);
	return 0;
}

编译运行输出：

frank@under:~/tmp$ gcc 259.c && ./a.out
0x765432EF

2.60

#include <stdio.h>

unsigned replace_byte(unsigned x, int i, unsigned char b);

int main(int argc, char const *argv[])
{
	printf("%#.8x\n", replace_byte(0x12345678, 2, 0xAB));
	printf("%#.8x\n", replace_byte(0x12345678, 0, 0xAB));
	return 0;
}

unsigned replace_byte(unsigned x, int i, unsigned char b)
{
	int move = i * 8;
	return x & ~(0xFF << move) | b << move;
}

编译运行输出：

frank@under:~/tmp$ gcc 260.c && ./a.out
0x12ab5678
0x123456ab

2.61

#include <stdio.h>

int main(int argc, char const *argv[])
{

	int x, y;	/* y means 0 should be returned */
	int sizeof_int = sizeof(int);
	
	/*condition A*/
	x = ~0;
	y = 0xFFFFFF00;
	printf("%d\t%d\n", !(~x), !(~y));
	
	/*condition B*/
	x = 0;
	y = 0x000000FF;
	printf("%d\t%d\n", !x, !y);
	
	/*condition C*/
	x = 0x000000FF;
	y = 0x0000000F;
	printf("%d\t%d\n", !((x ^ 0xFF)<<((sizeof_int-1)<<3)), !((y ^ 0xFF)<<((sizeof_int-1)<<3)));

	/*condition D*/
	x = 0x00FFFFFF;
	y = 0x0FFFFFFF;
	printf("%d\t%d\n", !(x >> ((sizeof_int-1) << 3)), !(y >> ((sizeof_int-1) << 3)));
	return 0;
}

编译运行输出：

frank@under:~/tmp$ gcc 261.c && ./a.out
1	0
1	0
1	0
1	0

2.62

#include <stdio.h>

int int_shifts_are_arithmetic(void);
int int_shifts_are_logic(void);

int main(int argc, char const *argv[])
{
	
	printf("%d\n", int_shifts_are_arithmetic());
	printf("%d\n", int_shifts_are_logic());
	return 0;
}

int int_shifts_are_arithmetic(void)
{
	int x = ~0;
	return x >> 1 == x;
}

int int_shifts_are_logic(void)
{
	unsigned x = ~0;
	return x >> 1 == x;
}

我这里由于没有不同字长/不同机器，就暂时用unsigned 代替了一下逻辑右移。

编译运行输出：

frank@under:~/tmp$ ./a.out 
1
0

2.63

#include <stdio.h>

unsigned srl(unsigned x, int k);
int sra(int x, int k); 

int main(int argc, char const *argv[])
{
	printf("%#.8x\n", srl(0x80000000, 8));
	printf("%#.8x\n", sra(0x80000000, 8));
	return 0;
}

unsigned srl(unsigned x, int k) 
{
	/* Perform shift arithmetically */
	unsigned xsra = (int) x >> k;
	/*思路是由k形成诸如0x00FFFFFF这样的掩码，与xsra进行与操作从而将高位置零*/
	unsigned w = sizeof(int) << 3;
	unsigned mask = ~(((1 << k)-1)<<(w-k)); 
  	/*(1 << k)-1能够获得低位连续为1，高位为0的掩码，但是其不能达到全1，于是继续向左移w-k然后取反*/
	return mask & xsra; 
}

int sra(int x, int k)
{
	/* Perform shift logically */
	int xsrl = (unsigned) x >> k;
	/*这个题目的关键点是判断符号位是否为1，通过test &= xsrl，test为零如果符号位为0，否则test不变（处于符号位位置*/
	unsigned w = sizeof(int) << 3;
	int test = 1 << (w-1-k);
	test &= xsrl;
	int mask = ~(test - 1);
  	/*test为零时，~(test - 1)为全零，不会改变xsrl*/
	return mask | xsrl;
}

这个题目卡了一会，主要是不能用右移比较麻烦。

编译运行输出：

frank@under:~/tmp$ gcc 263.c && ./a.out
0x00800000
0xff800000

2.64

#include <stdio.h>

int any_odd_one(unsigned x);

int main(int argc, char const *argv[])
{
	printf("%d\t%d\n", any_odd_one(1011), any_odd_one(1024));
	return 0;
}

int any_odd_one(unsigned x)
{
	unsigned sizeof_unsigned = sizeof(unsigned);
	unsigned w = sizeof_unsigned << 3;
	return !!(x << (w-1));
}

编译运行输出：

frank@under:~/tmp$ gcc 264.c && ./a.out
1	0

2.65 （终于碰见个四星的。。。）

/*二分法/加法无法达到要求
第一次尝试：
int odd_ones(unsigned x);

int main(int argc, char const *argv[])
{
	int sizeof_int = sizeof(int);
	return 0;
}

int odd_ones(unsigned x)
{
	int mask1 = 0x55555555;
	int mask2 = 0x33333333;
	int mask3 = 0x0F0F0F0F;
	int mask4 = 0x00FF00FF;
	int mask_odd_or_even = 1

	x = ((x >> 1)& mask1) + (x & mask1);
	x = ((x >> 2)& mask2) + (x & mask2);
	x = ((x >> 4)& mask3) + (x & mask3);
	x = ((x >> 8)& mask4) + (x & mask4);
	x = (x >> 16) + x;
	return x & mask_odd_or_even;
}
第二次尝试：
int odd_ones(unsigned x);

int main(int argc, char const *argv[])
{
	unsigned x1 = 0xFF00FF00;
	unsigned x2 = 0xFF01FF00;
	printf("%d\t%d\n", odd_ones(x2), odd_ones(x1));
	return 0;
}

int odd_ones(unsigned x)
{
	int mask1 = 0x55555555;
	int mask2 = 0x33333333;
	int mask3 = 0x0F0F0F0F;
	int mask4 = 0x00FF00FF;
	int mask_odd_or_even = 1;

	x = ((x >> 1)& mask1) + (x & mask1);
	x = x & mask1;
	x = (x >> 16) + x;
	x = x & mask1;
	x = (x >> 8) + x;
	x = x & mask1;
	x = (x >> 4) + x;
	x = x & mask1;
	x = (x >> 2) + x;
	return x & mask_odd_or_even;
}
*/
//第三次尝试：使用二分法/异或
#include <stdio.h>

int odd_ones(unsigned x);

int main(int argc, char const *argv[])
{
	unsigned x1 = 0xFF00FF00;
	unsigned x2 = 0xFF01FF00;
	printf("%d\t%d\n", odd_ones(x2), odd_ones(x1));
	return 0;
}

int odd_ones(unsigned x)
{
	x = x ^ (x >> 16);
	x = x ^ (x >> 8);
	x = x ^ (x >> 4);
	x = x ^ (x >> 2);
	x = x ^ (x >> 1);
	return x & 1;
}

~~这个题的关键点在于如何表示偶数（将所有“1”相加末位为0）以及类似二分法的相加方法，同时注意到每次需要用用掩码将以前的高位置零。~~

~~这个题目是不会有“溢出”的情况的，因为1+1=10,10+10=0100,0100+0100=00001000......datalab实验里有一个相似的题目，那个题目更难一些。~~

以上想法在满足“Your code should contain a total of at most 12 arithmetic, bitwise, and logical
operations.”时出现了问题，根本原因在于二分法需要顾及到低位相加可能产生的进位，所以每次都需要用掩码将特定的高位置零，思路有点受到之前datalab实验的束缚（那个是要计算“1”的总数目）。这里计算的是“1”的数目的奇偶，不用考虑进位，异或运算是最佳选择，因为1+1和0+0均产生0（代表偶数），1+0和0+1均产生1（代表奇数）。

编译运行输出：

frank@under:~/tmp$ gcc 265.c && ./a.out 
1	0

2.66

#include <stdio.h>
#include <limits.h>

/*
 * Generate mask indicating leftmost 1 in x. Assume w=32.
 * For example, 0xFF00 -> Ox8000, and Ox6600 --> Ox4000.
 * If x = 0, then return 0.
 */

int leftmost_one(unsigned x);

int main(int argc, char const *argv[])
{
	printf("%#.8x\n", leftmost_one(0xFF00));
	printf("%#.8x\n", leftmost_one(0x6600));
	printf("%#.8x\n", leftmost_one(0x88886600));
	printf("%#.8x\n", leftmost_one(0));
	return 0;
}

int leftmost_one(unsigned x)
{
	unsigned sizeof_unsigned = sizeof(unsigned);
	unsigned w = sizeof_unsigned << 3;
	x |= x >> 1;
	x |= x >> 2;
	x |= x >> 4;
	x |= x >> 8;
	x |= x >> 16;
	return x & ((~x >> 1)|INT_MIN);
}
/*
 * Your code should contain a total of at most 15 arithmetic, bitwise, and logical
 * operations.
 * Hint: First transform x into a bit vector of the form [O · · · 011 · . · 1].
 */

最后与INT_MIN做或运算是为了处理0x80000000这种边界情况，在这种情况下，~x >> 1由于没有更高位，而ｘ又是unsigned类型，所以最高位会是０而非１，为了适应这种情况，强制将~x >> 1最高位置１。

编译运行输出：

frank@under:~/tmp$ gcc 266.c && ./a.out 
0x00008000
0x00004000
0x80000000
00000000

另外，Web Asides http://csapp.cs.cmu.edu/3e/waside/waside-tneg.pdf 上面有一个利用-x和x的区别在于除最右１之前位翻转的特性求rightmost_one:　x&-x,　有时间可以看看。

2.67

(C11, 6.5.7p3) "If the value of the right operand is negative or is greater than or equal to the width of the promoted left operand, the behavior is undefined"

#include <stdio.h>

int int_size_is_32();

int main(int argc, char const *argv[])
{
	printf("%d\n", int_size_is_32());
	return 0;
}

int int_size_is_32()
{
	int set_msb = 1 << 31;
	int beyond_msb = set_msb;
	beyond_msb <<= 1;
	return set_msb && !beyond_msb;
}

编译运行输出：

frank@under:~/tmp$ gcc 267.c && ./a.out 
1

#include <stdio.h>

int int_size_is_32();

int main(int argc, char const *argv[])
{
	printf("%d\n", int_size_is_32());
	return 0;
}

int int_size_is_32()
{
	int set_msb = 1 << 15;
	set_msb <<= 15;
	set_msb <<= 1;
	int beyond_msb = set_msb;
	beyond_msb <<= 1;
	return set_msb && !beyond_msb;
}

2.68

#include <stdio.h>

int lower_one_mask(int n);

int main(int argc, char const *argv[])
{
	printf("%#.8x\n", lower_one_mask(6));
	printf("%#.8x\n", lower_one_mask(17));
	return 0;
}

int lower_one_mask(int n)
{
	int sizeof_int = sizeof(int);
	unsigned x = ~0;
	x >>= ((sizeof_int << 3) - n);
	return x;
}

编译运行输出：

frank@under:~/tmp$ gcc 268.c && ./a.out
0x0000003f
0x0001ffff

2.69

#include <stdio.h>

unsigned rotate_left(unsigned x, int n);

int main(int argc, char const *argv[])
{
	unsigned x = 0x12345678;
	printf("%#.8x\n", rotate_left(x, 0));
	printf("%#.8x\n", rotate_left(x, 4));
	printf("%#.8x\n", rotate_left(x, 20));
	return 0;
}

unsigned rotate_left(unsigned x, int n)
{
	unsigned sizeof_unsigned = sizeof(unsigned);
	unsigned w = sizeof_unsigned << 3;
	unsigned mask = ((1 << n)-1) << (w-n);
	unsigned cache = (mask & x) >> (w-n);
	x <<= n;
	return x | cache;
}

关键点在于掩码的产生和移除位数据的保存。

编译运行输出：

frank@under:~/tmp$ gcc 269.c && ./a.out 
0x12345678
0x23456781
0x67812345

2.70

#include <stdio.h>
#include <limits.h>

int fits_bits(int x, int n);

int main(int argc, char const *argv[])
{
    /*test short and 31bits*/
	printf("%d\n", fits_bits(-32768, 16));
	printf("%d\n", fits_bits(32767, 16));
	printf("%d\n", fits_bits(INT_MAX, 32));
	printf("%d\n", fits_bits(INT_MIN, 32));
	printf("%d\n", fits_bits(0, 16));
	printf("%d\n", fits_bits(0, 32));
	printf("%d\n", fits_bits(32768, 16));
	printf("%d\n", fits_bits(-32769, 16));
	printf("%d\n", fits_bits(INT_MIN, 31));
	printf("%d\n", fits_bits(INT_MAX, 31));
	return 0;
}

int fits_bits(int x, int n)
{
	unsigned sizeof_int = sizeof(int);
	unsigned w = sizeof_int << 3;
	int y = x << (w-n) >> (w-n);
	return y == x;
}

编译运行输出：

frank@under:~/tmp$ gcc 270.c && ./a.out 
1
1
1
1
1
1
0
0
0
0

2.71

/* Declaration of data type where 4 bytes are packed into an unsigned */
typedef unsigned packed_t;
/* Extract byte from word. Return as signed integer */
int xbytte(packed_t word, int bytenum);
/*That is, the function will extract the designated byte and sign extend it to be
a 32-bit int.
Your predecessor (who was fired for incompetence) wrote the following code:*/

//Failed attempt at xbyte:

int xbyte(packed_t word, int bytenum)
{
return (word>> (bytenum << 3)) & OxFF;
}
//A. What is wrong with this code?

//B. Give a correct implementation of the function that uses only left and right
//shifts, along with one subtraction.

当取出的字节为负数时，由于原操作“粗暴”的将高位置零，会返回一个错误的正值。

#include <stdio.h>

typedef unsigned packed_t;

int xbytte(packed_t word, int bytenum);

int main(int argc, char const *argv[])
{
	packed_t word = 0x8008FF00;
	printf("%d\n", xbytte(word, 0));
	printf("%d\n", xbytte(word, 1));
	printf("%d\n", xbytte(word, 2));
	printf("%d\n", xbytte(word, 3));
	return 0;
}

int xbytte(packed_t word, int bytenum)
{
	unsigned left_move = (3 - bytenum) << 3;
	unsigned right_move = (3) << 3;
	return (int)word << left_move >> right_move;
}

编译运行输出：

frank@under:~/tmp$ gcc 271.c && ./a.out 
0
-1
8
-128

2.72

/*BUGGY: Copy integer into buffer if space is available */
void copy_int(int val; void *buf, int maxbytes) 
{
	if (maxbytes-sizeof(val) >= 0)
		memcpy(buf, (void*) &val, sizeof(val));
}

sizeof返回的类型为size_t：

According to the 1999 ISO C standard (C99), size_t is an unsigned integer type of at least 16 bit (see sections 7.17 and 7.18.3).

size_tis an unsigned data type defined by several C/C++ standards, e.g. the C99 ISO/IEC 9899 standard, that is defined in stddef.h.1 It can be further imported by inclusion ofstdlib.h as this file internally sub includes stddef.h.

所以maxbytes-sizeof(val)将一直转化为无符号数并永远大于等于零。

void copy_int(int val; void *buf, int maxbytes) 
{
  	if(maxbytes < 0)
      	return;
	if (maxbytes >= sizeof(val))
		memcpy(buf, (void*) &val, sizeof(val));
}

2.73

#include <stdio.h>
#include <limits.h>

int saturating_add(int x, int y);

int main(int argc, char const *argv[])
{
	printf("%d\n", saturating_add(123456, -54321));
	printf("%d\n", saturating_add(2147483647, 1));
	printf("%d\n", saturating_add(-2147483648, -1));
	return 0;
}

int saturating_add(int x, int y)
{
	unsigned sizeof_int = sizeof(int);
	unsigned w = sizeof_int << 8;
	int i = (x ^ y) >> (w-1);//+-:FFFFFFFF ++/--:00000000
	int j = ((x+y) ^ x) >> (w-1);//overflow:FFFFFFFF otherwise:00000000
	int k = x >> (w-1);//+:00000000 -:FFFFFFFF
	return (i & (x + y)) + (~i & (j & ( (~k & INT_MAX) + (k & INT_MIN) )));
}

解释一下i j k：这三个变量和与运算结合用来做“判断语句”，i通过x，y是否异号判断是否可能溢出，j通过结果和加数的符号判断在同号的情况下是否发生溢出。k判断是应该返回INT_MAX 还是 INT_MIN。

编译运行输出：

frank@under:~/tmp$ gcc 273.c && ./a.out 
69135
2147483647
-2147483648

2.74

#include <stdio.h>
#include <limits.h>

int tsub_ok(int x, int y);

int main(int argc, char const *argv[])
{
	printf("%d\n", tsub_ok(123456, 54321));
	printf("%d\n", tsub_ok(2147483647, -1));
	printf("%d\n", tsub_ok(-2147483648, 1));
	return 0;
}

int tsub_ok(int x, int y)
{
	unsigned sizeof_int = sizeof(int);
	unsigned w = sizeof_int << 8;
	y = -y;

	int i = (x ^ y) >> (w-1);//+-:FFFFFFFF ++/--:00000000
	int j = ((x+y) ^ x) >> (w-1);//overflow:FFFFFFFF otherwise:00000000
	return i || ~j;
}

原理与2.73类似。

编译运行输出：

frank@under:~/tmp$ gcc 274.c && ./a.out 
1
0
0

2.75

#include <stdio.h>
#include <stdint.h>
#include <stdbool.h>

unsigned unsigned_high_prod(unsigned x, unsigned y);
int signed_high_prod(int x, int y);

int main(int argc, char const *argv[])
{
	/* code */
	return 0;
}

unsigned unsigned_high_prod(unsigned x, unsigned y)
{
	unsigned w = sizeof(int32_t) << 3;
	int64_t signed_total_prod = signed_high_prod(x, y);
	signed_total_prod <<= w;
	signed_total_prod += x*y;
	bool x_w = x < 0 ? true : false;
	bool y_w = y < 0 ? true : false;
	int64_t unsigned_total_prod = signed_total_prod + ((x_w*(int)y + y_w*(int)x)<<w) + x_w*y_w<<(w*2);
	return (unsigned)(unsigned_total_prod>>w);	
}

原理参见树上2.18等式。

2.76

void *calloc(size_t nmemb, size_t size)
{
	void *p;
	if(!(nmemb*size) || !(p = malloc(size*nmemb)))
		return NULL;
	else if (((size_t)(nmemb*size))/size != nmemb) 
                /* __builtin_umull_overflow() works too */
	{
                /* Thanks to zhzhwz who found a forget-to-free problem here. 
                   Maybe we should write a free_and_return_NULL block 
                   and goto it specifically.
                */
		fprintf(stderr, "size*nmemb overflow size_t.\n");
                free(p);
                return NULL;
	}
	else
	{
		memset(p, 0, size*nmemb);
		return p;
	}
}

2.77

#include <stdio.h>

int main(int argc, char const *argv[])
{
	unsigned sizeof_int = sizeof(int);
	unsigned w = sizeof_int << 3;
	int x = 1;
	printf("%d\n", (x << 4) + x);//k=17
	printf("%d\n", x - (x << 3));//k=-7
	printf("%d\n", (x << 6) - (x << 2));//k=60
	printf("%d\n", (x << 4) - (x << 7));//k=-112
	return 0;
}

编译运行输出：

frank@under:~/tmp$ gcc 277.c && ./a.out 
17
-7
60
-112

2.78

#include <stdio.h>

int divide_power2(int x, int k);

int main(int argc, char const *argv[])
{
	printf("%d\n", divide_power2(1024, 2));
	printf("%d\n", divide_power2(5, 2));
	printf("%d\n", divide_power2(-1024, 2));
	printf("%d\n", divide_power2(-5, 2));
	return 0;
}

int divide_power2(int x, int k)
{
	int bias = (1 << k) - 1;
	unsigned sizeof_int = sizeof(int);
	unsigned w = sizeof_int << 3;
	int judge = x >> (w-1); //-:FFFFFFFF +:00000000
	return (judge & ((x + bias) >> k)) + (~judge & (x >> k));
}

编译运行输出：

frank@under:~/tmp$ gcc 278.c && ./a.out 
256
1
-256
-1

2.79

int mul3div4(int x)
{
	int k = 2;
	int bias = (1 << k) - 1;
	unsigned sizeof_int = sizeof(int);
	unsigned w = sizeof_int << 3;
	x = (x << 1) + x;
	int judge = x >> (w-1); //-:FFFFFFFF +:00000000
	return (judge & ((x + bias) >> k)) + (~judge & (x >> k));
}

2.80

[BUG] threefourths(3)应计算得到2，代码将会得到0，可以参考 2.80
From: zhzhwz

int threefourths(int x)
{
  	int k = 2;
	int bias = (1 << k) - 1;
	unsigned sizeof_int = sizeof(int);
	unsigned w = sizeof_int << 3;
	int judge = x >> (w-1); //-:FFFFFFFF +:00000000
	x = (judge & ((x + bias) >> k)) + (~judge & (x >> k));
  	return (x << 1) + x;
}

参考2.78

2.81

x = ~0 << k;

x = ~(~0 << k) << j;

2.82

A: when x = INT_MIN, y = 0. Yields 0

B: It always yields 1. Say, "mod" = mod 2^32. LEFT = ((((((x+y)mod)*16)mod)+y)mod)-x)mod = (17y + 15x)mod. RIGHT = (17*y)mod + (15*x)mod = (17y + 15x)mod.

C: It always yields 1. LEFT = (~x + 1) + (~y + 1) - 1 = -x + -y - 1. RIGHT = ~(x+y) + 1 - 1 = -(x+y) - 1 = -x + -y - 1.

D: It always yields 1. Since whether a integer type data is a int or unsigned doesn't influence the implementations of subtraction or unary minus operators.

（写到一半突然发现是用英文写的，可能是看教材影响的。。。）

E: 永远产生1，因为x先向右移动，再向左移动相同的位置，所以数据的高位不会受到影响。但是如果低两位有1的话，会都变为0。因为低位的1在补码中无论对于负数还是正数都是加的。所以对于正数来说，值会变小或者不变；对于0来说，值会不变；对于负数来说，值会变小或者不变。

2.83

根据提示：Y = x*2^k - x 即 x = Y/(2^k - 1)

由A：(a) Y = 101 = 5, k = 3, x = 5/7 (b) Y = 0110 = 6, k = 4, x = 2/5 (c) Y = 010011 = 19, k = 6, x = 19/63

2.84

#include <stdio.h>

int float_le(float x, float y);
unsigned f2u(float x);

int main(int argc, char const *argv[])
{
	printf("%d\n", float_le((float)1.11, (float)1.10));
	printf("%d\n", float_le((float)-1.2, (float)3.0));
	printf("%d\n", float_le((float)1.3, (float)1.3));
	printf("%d\n", float_le((float)0, (float)0));
	printf("%d\n", float_le((float)-1.1, (float)0));
	printf("%d\n", float_le((float)0, (float)1.1));
	return 0;
}

unsigned f2u(float x)
{
	return *(unsigned*)&x;
}

int float_le(float x, float y) 
{
	unsigned ux = f2u(x);
	unsigned uy = f2u(y);
	/*Get the sign bits*/
	unsigned sx = ux >> 31; //+:0 -:1
	unsigned sy = uy >> 31;

	/* Give an expression using only ux, uy, sx, and sy */
	return (sx ^ sy) ? (sx ? 1 : 0)/*-+ +-*/ : (sx ? (ux>=uy) : (ux<=uy))/*-- ++*/;
}

编译运行输出：

frank@under:~/tmp$ gcc 284.c && ./a.out 
0
1
1
1
1
1

[PATCH] +0和-0不等
From: zhzhwz

@@ -11,6 +11,7 @@ int main(int argc, char const *argv[])
printf("%d\n", float_le((float)0, (float)0));
printf("%d\n", float_le((float)-1.1, (float)0));
printf("%d\n", float_le((float)0, (float)1.1));
    printf("%d\n", float_le((float)(1/1e100), (float)(-1/1e100)));
return 0;
}
@@ -28,5 +29,5 @@ int float_le(float x, float y)
unsigned sy = uy >> 31;
/* Give an expression using only ux, uy, sx, and sy */
    return (sx ^ sy) ? (sx ? 1 : 0)/-+ +-/ : (sx ? (ux>=uy) : (ux<=uy))/-- ++/;
    return (ux << 1 == 0 && uy << 1 == 0)/all 0/ || ((sx ^ sy) ? (sx ? 1 : 0)/-+ +-/ : (sx ? (ux>=uy) : (ux<=uy))/-- ++/);
}
\ No newline at end of file

2.85

bias = 2^(k-1) - 1 suppose that k <= n

E = 0b10+bias, M = 0b1.11, f = ob1100*, V = 1.0

bit representation: 0, 0b10+bias, 1100*

E = n+bias, M = 0b1.11*, f = 0b11*, V = 2^(n+1)-1

bit representation: 0, n+bias, 11*

The smallest positive normalized value : E = 0b00*1, M = 0b1.00*, f = 0b00*, V = 1.0

So the reciprocal is exactly the same number.

2.86

bias = 2^14 - 1

Smallest positive denormalized:

Value: 0, 00*, 0, 00*1 Decimal:2^(2-214) * 2^(-63)

Smallest positive normalized:

Value: 0, 00*1, 1, 00* Decimal:2^(2-214)

Largest normalized:

Value: 0, 11*0, 1, 11* Decimal: 2⁽²14-1) * (2-2^(-63))

2.87和2.88本来在Typora上是用表格写的，上传上来好像有格式问题，将就看一下 ; )

2.87

| Description | Hex | M | E | V | D |
| -0 | 8000 | 0 | -14 | 0 | 0 |
| Smallest value > 2 | 4001 | 1025/1024 | 1 | 10252^-8 | 2.001953 |
| 512 | 6000 | 1 | 9 | 12^9 | 512.000000 |
| Largest denormalized | 0311 | 1023/1024 | -14 | 10232^-24 | 0.000061 |
| negative infinite | FC00 | - | - | - | -inf |
| 3BB0 | 3BB0 | 124/64 | -1 | 312^-5 | 0.968750 |

2.88

| Format A | Format A | Format B | Format B |
| Bits | Value | Bits | Value |
| 1 01111 001 | -9/8 | 1 0111 0010 | -9/8 |
| 0 10110 011 | 112^4 | 0 1110 0110 | 112^4 |
| 1 00111 010 | -52^-10 | 1 0000 0101 | -52^-10 |
| 0 00000 111 | 72^-17 | 0 0000 0001 | 2^-10 |
| 1 11100 000 | -2^13 | 1 1110 1111 | -312^3 |
| 0 10111 100 | 32^7 | 0 1110 1111 | 312^3 |

2.89

A：总是返回１．因为int到double不会有精度上的损失，所以ｘ，dx转float(损失精度)的结果是一样的。

B：不总是返回１．如x=INT_MIN，Y=1。

[PATCH] 注意2.89的C中dz是由一个整数转换来的，因此不会取到1e-30
From: zhzhwz

- C：不总是返回１．浮点数不满足结合律，如dx=1e30, dy=-1e30, dz=1e-30。
+ C：由于int转换为double不会有精度上的损失，且在int范围内使用double做加法得到的结果一定是精确的（这是因为double的尾数足以容纳32bit），因此满足结合律。

D：不总是返回１．原因同上。例如dx与dy互为倒数且dy*dz=+infinite。

E：不总是返回１．例如dx=1.0, dz=0.0　。

2.90

float fpwr2(int x)
{
	/* Result exponent and fraction */
	unsigned exp, frac;
	unsigned u;
	if (x < -149)
	{
		/* Too small. Return 0.0 */
		exp = 0;
		frac = 0;
	} 
	else if (x < -126) 
	{
		/* Denormalized result */
		exp = 0;
		frac = 1 << (149 + x);
	}
	else if (x < 128)
	{
		/* Normalized result. */
		exp = x + 127;
		frac = 0;
	}
	else
	{
		/* Too big. Return +oo */
		exp = 0xFF;
		frac = 0; 
	}
	
	/*Pack exp and frac into,32 bits */
	u = exp << 23 | frac;
	/* Return as float */
	return u2f(u);
}

2.91

0x 40490FDB = 0b 0100 0000 0100 1001 0000 1111 1101 1011 = 0,10000000,10010010000111111011011

A: 10010010000111111011011

B: (详情可见)2.83 y=1, k=3, 即　0b11.(001)*

C: 0x4049039b 0x40492492 从高位向低位第１９个。

浮点数部分由于时间所限，没有进行相关测试，思路大致应该是对的，可能会有一些边界／特殊情况会产生问题，欢迎指出。

2.92

float_bits float_negate(float_bits f)
{
	unsigned sign = f >> 31;
	unsigned exp = f >> 23 & 0xFF;
	unsigned frac = f & 0x7FFFFF;
	if(!(exp ^ 0xFF) && frac)
	{
		return f;
	}
	else
	{
		sign = !sign;
		return (sign << 31) | (exp << 23) | frac;
	}
}

2.93

float_bits float_absval(float_bits f)
{
	unsigned exp = f >> 23 & 0xFF;
	unsigned frac = f & 0x7FFFFF;
	if(!(exp ^ 0xFF) && frac)
	{
		return f;
	}
	return (exp << 23) | frac;
}

2.94

float_bits float_twice(float_bits f)
{
	unsigned sign = f >> 31;
	unsigned exp = f >> 23 & 0xFF;
	unsigned frac = f & 0x7FFFFF;
	if(!(exp ^ 0xFF))
	{
		if (frac)//NaN
		{
			return f;
		}
		/*else
		{
			return (sign << 31) | (exp << 23) | frac;//infinite
		}*/
	}
	else//Denormalnized and normalized
	{
		if (exp)//Normalnized
		{
			if (!(frac >> 22))
			{
				frac <<= 1;
				//return (sign << 31) | (exp << 23) | frac;
			}
			else if (!(exp ^ 0xFE))
			{
				++exp;
				//return (sign << 31) | (exp << 23) | frac;
			}
			else//overflow
			{
				frac = 0;
				exp = 0xFF;
				//return (sign << 31) | (exp << 23) | frac;
			}
		}
		else//Denormalized
		{
			if (!(frac >> 22))
			{
				frac <<= 1;
				//return (sign << 31) | (exp << 23) | frac;
			}
			else//Turn to Normalized
			{
				++exp;
				frac = frac << 10 >> 9;//set the 23th bit of frac to 0 and then left shift one bit.（注释是必要的。。。过了几天看这一段的时候自己也没弄懂，忘了这里frac是一个unsigned。。。） 
				//return (sign << 31) | (exp << 23) | frac;
			}
		}
	}
	return (sign << 31) | (exp << 23) | frac;
}

2.95

float_bits float_half(float_bits f)
{
	unsigned sign = f >> 31;
	unsigned exp = f >> 23 & 0xFF;
	unsigned frac = f & 0x7FFFFF;
	if(!(exp ^ 0xFF))
	{
		if (frac)//NaN
		{
			return f;
		}
		/*else
		{
			return (sign << 31) | (exp << 23) | frac;//infinite
		}*/
	}
	else//Denormalnized and normalized
	{
		if (exp)//Normalnized
		{
			if (exp != 1)
			{
				--exp;
				//return (sign << 31) | (exp << 23) | frac;
			}
			else//Turn to Denormalnized
			{
				if (frac)//maybe need to round to even
				{
					if ((frac >> 1)&1)
					{
						++frac;
						frac >>= 1;
						frac |= 0x400000;
						--exp;
					}
					else
					{
						frac >>=1;
						frac |= 0x400000;
						--exp;
					}
				}
			}
		}
		else//Denormalized
		{
			if (frac)//maybe need to round to even
			{
				if ((frac >> 1)&1)
				{
                                        ++frac;
					frac >>= 1;
				}
				else
				{
					frac >>= 1;
				}
			}
		}
	}
	return (sign << 31) | (exp << 23) | frac;
}

2.96

int float_f2i(float_bits f)
{
	unsigned sign = f >> 31;
	unsigned exp = f >> 23 & 0xFF;
	unsigned frac = f & 0x7FFFFF;
	unsigned bias = 127;
	int flag = 0;
	if(sign)//<=0
	{
		if (f < 0xBF800000)//>-1
		{
			return 0;
		}
		else if (f <= 0xCF000000)
		{
			if (f == 0xCF000000)//INT_MIN
			{
				return INT_MIN;//0x80000000;
			}
			else
			{
				f &= 0x7FFFFFFF;//first treat it as a positive number
				flag = 1;
				goto A;
				B:
				return (~f + 1);//-(+int)
			}
		}
		else//overflow/-infinite/NaN
		{
			return 0x80000000;
		}
	}
	else//>=0
	{
		if (f < 0x3F800000)//<1//Denormalnized->0
		{
			return 0;
		}
		else if (f <= 0x4EFFFFFF)
		{
			A:
			frac |= 0x800000;
			unsigned move = 23 - (exp-bias);
			if (flag)//jumped from a negative number
			{
				f = move >= 0 ? frac >> move : frac << -move;
				goto B;
			}
			else
			{
				return move >= 0 ? frac >> move : frac << -move; 
			} 
		}
		else//overflow/+infinite/NaN
		{
			return 0x80000000;
		}
	}
}

标准答案
 /* Compute (int) f. If conversion causes overflow or f is NaN, return 0x80000000 */
 int float_f2i(float_bits f) {
 unsigned sign = f >> 31;
 unsigned exp = (f >> 23) & 0xFF;
 unsigned frac = f & 0x7FFFFF;
 /* Create normalized value with leading one inserted, and rest of significand in bits 8--30./
 unsigned val = 0x80000000u + (frac << 8);
 if (exp < 127) {
 /* Absolute value is < 1 */
 return (int) 0;
 }
 if (exp > 158)
 /* Overflow */
 return (int) 0x80000000u;
 /* Shift val right */
 val = val >> (158 - exp);
 /* Check if out of range */
 if (sign) {
 /* Negative */
 return val > 0x80000000u ? (int) 0x80000000u : -(int) val;
 } else {
 /* Positive */
 return val > 0x7FFFFFFF ? (int) 0x80000000u : (int) val;
 }
 }

2.97

int leftmost_one(unsigned x)
{
	unsigned sizeof_unsigned = sizeof(unsigned);
	unsigned w = sizeof_unsigned << 3;
	x |= x >> 1;
	x |= x >> 2;
	x |= x >> 4;
	x |= x >> 8;
	x |= x >> 16;
	return x & ((~x >> 1)|INT_MIN);
}

float_bits float_i2f(int i)
{
	unsigned sign = 0;
	unsigned exp = 0;
	unsigned frac = 0;
	unsigned bias = 127;
	if (i == INT_MIN)
	{
		return 0xCF000000;
	}
	else//treat negative as positive
	{
		if (i < 0)
		{
			sign = 1;
			i = -i;
		}
		int mask = leftmost_one(i);
		int move = 0;
		if (mask >= 0x00800000)//rightshift
		{
			while(mask != 0x00800000)
			{
				mask >>= 1;
				++move;
			}

			if ((i & ((1 << (move+1)) - 1)) > (1 << move))//round to even(>1/2)
			{
				i >>= move;
				i += 1;
			}
			else if((i & ((1 << (move+1)) - 1)) < (1 << move))//(<1/2)
			{
				i >>= move;
			}
			else// 1/2
			{
				if ((i >> move)&1)//round to even
				{
					i >>= move;
					i += 1;
				}
				else
				{
					i >>= move;
				}
			}
		}
		else//leftshift
		{
			while(mask != 0x00800000)
			{
				mask <<= 1;
				--move;
			}
			i <<= -move;
		}
		frac = i & 0x7FFFFF;//Discard the 24th bit one
		exp = bias + 22 + move;
	}
	return (sign << 31) | (exp << 23) | frac;
}

用到了2.66产生标志整数最高位１掩码，从而判断应该左移或者右移多少位。

posted @ 2017-09-26 14:13 QiuhaoLi 阅读(12039) 评论(6) 编辑收藏举报

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深入理解计算机系统_3e 第二章家庭作业 CS:APP3e chapter 2 homework

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