十进制转二进制错误

Question

I am trying to create a program that converts decimal to binary and prints 0b before the binary.

I need the program to print 0b0 if 0 is the input, and 0b11101 when input is 29. I can seem to do one, but not the other, and when I fix that bug, the other stops working

Where am I going wrong ?

#include <stdio.h>
int main()
{

int n,i = 0,j=0,k;

scanf("%d",&n);
int bin[100];

    k=n;

    //printf("n is %d \n",n);
    while(n!=0)
    {
        //printf("n is %d :\n",n);
     bin[i]= n%2;
        //printf ("n/2 is  %d\n",n%2);
     n=(n-n%2)/2;
     i++;
        //printf("i is %d\n",i);
    }
printf("0b");
    //if (n='0')
    //{
    //  printf("0");
    //}


    //else
    //{
        for (j=i;j>0;j--)
        {
            //printf("\n j is,%d and there ,is %d\n",j,bin[(j-1)]);
            printf("%d",bin[(j-1)]);

        }
    //}

}

Answer 1

0 does not need to be treated as a special case. Simply fix: use a do loop rather than while () .

#include <stdio.h>
int main() {
  int n,i = 0,j=0,k;
  scanf("%d",&n);
  int bin[100];
  // k=n;  // not needed

  // while(n!=0) {
  do {
    bin[i]= n%2;

    // Another simplification
    // n=(n-n%2)/2;
    n /= 2;

    i++;
  } while (n);
  printf("0b");
  for (j=i;j>0;j--) {
    printf("%d",bin[(j-1)]);
  }
}

Other coding needed to handle n<0 .

Answer 2

#include <stdio.h>
int main()
{
int n,i = 0,j=0,k;
scanf("%d",&n);
int bin[100]; 
k=n;
    while(n!=0)
    {
     bin[i]= n%2;
     n=(n-n%2)/2;
     i++;
    }
printf("0b");
    if (k==0)
    {
      printf("0");
    }
    else
    {
        for (j=i-1;j>=0;j--)
        {
            printf("%d",bin[(j-1)]);
        }
    }
}

Answer 3

A naive and more portable implementation of a decimal to binary conversion function ...

#include <limits.h>

void print_binary(unsigned x){
    unsigned n=UINT_MAX;

    n = ~(n >> 1);

    while (n){ 
        if (n & x)
            putchar('1');
        else
            putchar('0');
        n = n >> 1;
    }
    putchar('\n');
}