Question

int main()
{
   int n = 56;
   char buf[10]; //want char* buf instead
   sprintf(buf, "%i", n);
   printf("%s\n", buf);

   return 0;
}

This piece of code works, my question is what if i want the same behaviour, with buf being a char* ?

Answer 1

buf is an array statically allocated on the stack. For it to be of type char * you need to allocate dynamically the memory. I guess what you want is something like this:

int main()
{
   int n = 56;
   char * buf = NULL;
   buf = malloc(10 * sizeof(char));
   sprintf(buf, "%i", n);
   printf("%s\n", buf);

   // don't forget to free the allocated memory
   free(buf);

   return 0;
}

EDIT: As pointed out by ‘Thomas Padron-McCarthy’ in the comments, you should not cast the return of malloc(). You will find more info here. You could also remove completely the sizeof(char) as it will always by 1.