How to append null terminator to end of the indexed char pointer

Question

I ran the following code and it crashes with the while loop running forever. When I debugged this code, I found the problem at *(pointer+cnt)='\0'; the null character is never there. I don't know how to append the null terminator here so that the program doesn't crash.

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

char* decimal_binary(int);

int main()
{
   int n;
   char *ptr=NULL;

   printf("Enter the number\n");
   scanf("%d",&n);

   ptr=decimal_binary(n);
   //printing out the characters
   while(ptr!='\0')
   {
     printf("%c",*ptr);
     ptr++;
   }
   free(ptr);
   return 0;
}

char* decimal_binary(int n)
{
  int c,d,cnt=0;
  char *pointer=(char*)malloc(8+1);
  if(pointer==NULL)
     exit(EXIT_FAILURE);

  for(c=7;c>=0;c--)
  {
    d=n>>c;
    if(d&1)
        *(pointer+cnt)=1+'0';
    else
        *(pointer+cnt)=0+'0';
    cnt++;
   }
//Null not getting added at the end of this sequence.Hence while loop in    main runs forever.
*(pointer+cnt)='\0';
return pointer;
}

Answer 1

You've chosen to write:

while(ptr!='\0')

That's a funny way of writing:

while (ptr != 0)

or:

while (ptr != NULL)

where you intended to write:

while (*ptr != '\0')

The conventional way of writing *(pointer+cnt) is pointer[cnt].

You can't afford to free the incremented pointer; you must free what was returned by malloc() — or calloc() or realloc() or …

Preserve a copy of the value returned by binary_decimal() and free the copy (or increment the copy and free the value in ptr).

You can use either of the two binary_decimal() functions in the code below:

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

char *decimal_binary(int);

int main(void)
{
    int n;
    char *ptr = NULL;

    printf("Enter the number\n");
    scanf("%d", &n);

    ptr = decimal_binary(n);
    char *cpy = ptr;
    //printing out the characters
    while (*ptr != '\0')
    {
        printf("%c", *ptr);
        ptr++;
    }
    putchar('\n');
    free(cpy);
    return 0;
}

char *decimal_binary(int n)
{
    int cnt = 0;
    char *pointer = (char *)malloc(8 + 1);
    if (pointer == NULL)
        exit(EXIT_FAILURE);

    for (int c = 7; c >= 0; c--)
    {
        int d = n >> c;
        if (d & 1)
            pointer[cnt] = 1 + '0';
        else
            pointer[cnt] = 0 + '0';
        cnt++;
    }
    pointer[cnt] = '\0';
    return pointer;
}

Or:

char *decimal_binary(int n)
{
    int cnt = 0;
    char *pointer = (char *)malloc(8 + 1);
    if (pointer == NULL)
        exit(EXIT_FAILURE);

    for (int c = 7; c >= 0; c--)
        pointer[cnt++] = ((n >> c) & 1) + '0';
    pointer[cnt] = '\0';
    return pointer;
}

This could be compressed still more (and even less readably):

char *decimal_binary(int n)
{
    char *pointer = (char *)malloc(8 + 1);
    if (pointer == NULL)
        exit(EXIT_FAILURE);

    for (int c = 7; c >= 0; c--)
        pointer[7 - c] = ((n >> c) & 1) + '0';
    pointer[8] = '\0';
    return pointer;
}

And for a buffer of 9 bytes, you could perfectly well allocate a local variable in main() and pass the address to decimal_binary() so it does not need to use malloc() and the main() does not need to use free:

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

void decimal_binary(int, char *);

int main(void)
{
    int n;
    char buffer[9];
    char *ptr = buffer;

    printf("Enter the number\n");
    scanf("%d", &n);

    decimal_binary(n, buffer);

    while (*ptr != '\0')
    {
        printf("%c", *ptr);
        ptr++;
    }
    putchar('\n');

    return 0;
}

void decimal_binary(int n, char *pointer)
{
    for (int c = 7; c >= 0; c--)
        pointer[7 - c] = ((n >> c) & 1) + '0';
    pointer[8] = '\0';
}