IEEE-754 floating number binary represent

Question

I'm solving a question on online judge: https://acm.cs.nthu.edu.tw/problem/12237/

I need to represent IEEE-754 floating number with binary represent.

Here's my code:

#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);
    }
}

But I got few wrong answer and because of I didn't know the testcase , I can't find out if there's any exception lead to the wrong answer. Thanks for your help!

Make up your own test data, to the limits given in the problem. Watch out for corner cases. — Weather Vane, May 20 '20 at 13:24
The range of the floating point number is -10^20 ~ 10^20 , and I've tried them https://www.h-schmidt.net/FloatConverter/IEEE754.html and it's same as my program T_T — 王柏智, May 20 '20 at 13:48
When the input number is sub-normal, your code fails. You need a simpler way to convert to binary output, non-recursive (that's a "school exercise" method), for exactly 32 bits. — Weather Vane, May 20 '20 at 14:13
The binary output is faulty in other cases too, but it "happens to work" for the sample data given. Also you have a non-compliant function prototype `void binaryprint();` — Weather Vane, May 20 '20 at 14:27
! I found when my input is between 0~2 , it will only print 31bit and I need to print another "0" but I'm curious why — 王柏智, May 20 '20 at 14:35
As it happens, the smaller the number, the fewer bits it prints. Just iterate a mask along 32 bits, `for(unsigned m = 0x80000000; m != 0; m >>= 1)` Never use a recursive solution when you can use a simple iterative one. — Weather Vane, May 20 '20 at 14:37
Oh , I calculate one time myself , if x<2 the exponent will be 127(+0) , and can't fulfill 8 bits(at least 128) so I need to print another "0" :D — 王柏智, May 20 '20 at 14:49
The smallest number I tried only printed 2 digits :) Just use a different algorithm (and not one that suppresses leading or trailing zeros, for example). — Weather Vane, May 20 '20 at 14:51
OK!! Thank you very much! Actually I used recursive only because I didn't need to reverse it ( I think it is necessary to reverse the print if using for() ?) — 王柏智, May 20 '20 at 14:54

score 0 · Answer 1 · answered May 25 '20 at 18:34

Here is a some program, with explanations in comments, that handles the cases shown. (It does not handle infinities or NaNs.)

#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');
    }
}

IEEE-754 floating number binary represent

1 Answers1