Char concatenation and memory reallocation

Question

I am trying to concatenate single digits to a string:

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

int main() {

  char *test="";
  int i;

  for (i = 0; i < 10; i++) 
    char *ic;
    ic = malloc(2);
    sprintf(ic, "%d", i);

    // printf("%d\n", strlen(test));

    if (strlen(test) == 0) 
      test = malloc(strlen(ic));
      strcpy(test, ic);
    } else {
      realloc(test, strlen(test) + strlen(ic));
      strcat(test, ic);
    }

  }
  printf("%s\n", test);
  // printf("%c\n", test);
  free(test);
  test = NULL;
  return 0;
}

My goal is for the result of the final printf ("%s", test) be 0123456789.

Answer 1

Remember that strings are terminated with null characters. When you allocate memory for a string, you must add an extra byte for the null. So you'll need to add 1 to each of your malloc() and realloc() calls. For example:

test = malloc(strlen(ic) + 1);

Remember, also, that realloc() is allowed to "move" a variable to a new location in memory. It may need to do that in order to find enough contiguous unallocated space. It can also return NULL if it's unable to allocate the memory you've requested, so you should call it like this:

char *new_mem = realloc(test, strlen(test) + strlen(ic) + 1);
if (new_mem == NULL) {
  // Not enough memory; exit with an error message.
} else {
  test = new_mem;
}

Answer 2

A few issues:

1. char *test=""; - you point test to a constant C string. You don't write to it, but it's dangerous and will compile in C++. The type of "" is const char*.
2. strlen returns the length of the string not the buffer size. You need to add +1 to include the NULL character. This is your biggest issue.
3. A short known fixed size small buffer like ic should be allocated on the stack. A simple char array. You also forgot to free() it.