IEEE-754浮点数二进制表示

Question

我正在解决在线法官的问题：https://acm.cs.nthu.edu.tw/problem/12237/

我需要用二进制表示来表示IEEE-754浮点数。

这是我的代码：

#include <stdio.h>

void binaryprint();

int main(){
    float x;
    while(scanf("%f",&x) != EOF){                //File end with EOF
        unsigned int y = *(unsigned int*)&x;
        if(x == 0){                                  // INput = 0 -> 0*32
            for(int i = 0 ; i < 32 ; i++){
                printf("%d",0);
            }
        }
        else {
            if(x > 0) printf("%d",0);     //0 means positive but I find c will not print it out
            binaryprint(y);              //transfer to binary represent
        }
        printf("\n");
    }
}

void binaryprint(unsigned int x){
    if(x != 1){
        binaryprint(x/2);
        printf("%d",x%2);
    }
    else{
        printf("%d",1);
    }
}

但是我没有几个错误的答案，由于我不知道测试用例，我无法确定是否有任何例外导致错误的答案。感谢您的帮助！

Answer 1

这里是一些程序，带有注释的解释，可以处理所示的情况。 （它不处理无穷大或NaN。）

#include <math.h>
#include <stdio.h>

int main(void)
{
    float x;
    while (scanf("%f", &x) == 1)
    {
        //  Print the sign bit.
        putchar('0' + (x < 0));
        if (x < 0)
            x = -x;

        /*  Multiply or divide x by two to get a significand in the interval
            [1, 2).  At the same time, count the operations to figure what
            power of two (in Exponent) is needed to scale that significand to
            the original value of x.
        */
        int Exponent = 0;
        while (x < 1)
        {
            Exponent -= 1;
            //  If we reach the subnormal range, stop.
            if (Exponent < -126)
                break;
            x *= 2;
        }
        while (2 <= x)
        {
            Exponent += 1;
            x /= 2;
        }

        /*  Encode the exponent by adding the bias of the IEEE-754 binary32
            format.
        */
        Exponent += 127;

        //  Print the seven bits of the exponent encoding.
        for (int i = 7; 0 <= i; --i)
            putchar('0' + (Exponent >> i & 1));

        /*  Skip the leading bit of the significand and print the 23 trailing
            bits.
        */
        x -= (int) x;
        for (int i = 0; i < 23; ++i)
        {
            x *= 2;
            int bit = x;
            x -= bit;
            putchar('0' + bit);
        }

        putchar('\n');
    }
}