深入理解计算机系统 家庭作业 2.97

核心思路:按照书上的分,符号位,exp段,frac段处理.

舍入也是按书上的来,把舍弃的部分与中值相比较.

舍入之前必须先把frac和exp合并在一块 根据舍弃的情况加一  这样加一如果frac段的进位进到了exp段就会加到exp段

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

typedef unsigned float_bits;
float_bits float_i2f(int i);
unsigned bits_mask(int l);

int main() 
{
    int i= 2147483583;//这个数+1就会被舍入到INT_MAX;
    printf("i2f=%x\n",float_i2f(i));

	return 0;
}

unsigned bits_mask(int l) {
  return (unsigned) -1 >> (32-l);
}

float_bits float_i2f(int i)
{
    printf("i=%x\n",i);
    int n=0;
    int times=0;
    unsigned mask=0x40000000;
    unsigned sign = i >> 31;
    unsigned exp = 0;
    unsigned frac = 0;
    unsigned num= (unsigned)i;//因为后面涉及到右移
	if (i == 0) {
	    sign = 0;
	    exp = 0;
	    frac = 0;
    	return sign << 31 | exp << 23 | frac;
	}
	if (i == INT_MIN) {
	    sign = 1;
	    exp = 127 + 31;
	    frac = 0;
    	return sign << 31 | exp << 23 | frac;
	}
    if(sign){
		num=~num+1;
    }
    num = num & 0x7fffffff;
    //通过循环找到i的最高位.
    while(n==0){
	    n = num & mask;//从31位开始取位
	    times++;
	    mask>>=1;
	}
	unsigned leng = 32-times;//得到参数i的二进制位数
	exp = 127+leng-1;
	if(leng<=24){//整数不长了 浮点数的frac的23位可以容纳
		num = num<<(23-leng+1);
		num = (exp << 23) |(num&0xff7fffff);//隐含第24位得1
	}
	else{//整数太长了,超过frac的23位 
		int move = leng-1-23;
		unsigned round_part = num&bits_mask(move);
		unsigned round_mid = 1<<(move-1);
		num = (exp << 23) |((num>>move)&0xff7fffff);
	    if (round_part < round_mid) {
	    }
		else if (round_part > round_mid)
	      num += 1;
		else{
		    if ((num & 0x1) == 1)
		        num += 1;
		}
	}
	return sign << 31 | num;
}

//https://www.h-schmidt.net/FloatConverter/IEEE754.html IEEE-754浮点转换器 用于检查结果

写了好几天了,断断续续写 参考了2.97 - CASPP 3e Solutions,建议一定要自己写一遍这个,这一题和书本内容紧密相连.