在 C 中将十进制转换为具有数十亿数字的二进制

Question

#include<stdio.h>    
#include<stdlib.h>  

int main() {
    long long *a, n, i;
    while (0 != 1) {
        printf("Enter the number to convert: ");
        scanf("%lli", &n);
        a = (int*) malloc(n * sizeof (int));
        printf("%p", a);
        for (i = 0; n > 0; i++) {
            a[i] = n % 2;
            n = n / 2;
        }
        printf("\nBinary of Given Number is=");
        for (i = i - 1; i >= 0; i--) {
            printf("%lli", a[i]);
        }
        __fpurge(stdin);
        getchar();
    }
}

I have this code, my teacher request input number 77777777777777777777777777777777777777777777 and convert to binary. But program can't run with this number, it's too long. So anyone can help me, how to input billion number and program can run.

Answer 1

Since your input is too large for an int type you can use a string and manipulate it using the same algorithm.

#include <stdio.h>    
#include <stdlib.h>  
#include <string.h>
#include <stdbool.h>

bool isZero(const char *s, int len)
{
    for (int i = 0; i < len; ++i)
    {
        if (s[i] != '0')
        {
            return false;
        }
    }
    return true;
}

void div2(char *s, int len)
{
    int remainder = 0;
    for (int i = 0; i < len; ++i)
    {
        int digit = s[i] - '0' + remainder;
        remainder = 10 * (digit % 2);
        s[i] = '0' + (digit / 2);
    }
}

int main()
{
    char input[256] = "77777777777777777777777777777777777777777777";
    char bits[2048] = { 0 };

    int len = strlen(input);
    for (int i = 0; i < len; ++i)
    {
        if (input[i] < '0' || input[i] > '9')
        {
            printf("%c is not a numeric digit, exiting.\n", input[i]);
            return -1;
        }
    }

    int pos = 0;
    while (!isZero(input, len))
    {
        int digit = input[len - 1] - '0';
        bits[pos++] = '0' + (digit % 2);
        div2(input, len);
    }

    len = strlen(bits);
    while (len--)
    {
        putchar(bits[len]);
    }
    putchar('\n');

    return 0;
}