Question

I’m trying to hunt down memory leaks and have found one source. I am malloc’in the pointer in one function and freeing it in another, but I’m missing out on understanding how to copy the value the pointer points to while also being able to free the pointer.

Current implementation (with memory leak):

// This code has been greatly simplified
// and as such does not appear to have any purpose
int foo(){
  int bestval = 0;
  char *best;
  char *match;
  for (int i=0;i<3;i++) {
      int returnValue = bar(&best);
      if (returnValue > 10) {
        (1)
         match = best;
      }
  }

  printf("%s", match); 
  (2)    
  return 0;
}


int bar(char ** best) {
  char*s = "Hello!";
  *best = malloc(strlen(s) + 1);
  strcpy(*best,s);
  return 0;
}

Two questions

1. If I had to free memory at (1) rather than (2), how would I do it so that match would still have what was contained in best?

2. Should I be doing strcpy to copy best to match? If so, do I have to do another malloc within foo?

Answer 1

A bit of a stab in the dark, assuming there’s a loop in Foo…

int foo()
{   
    int bestval = 0;   
    char *best;   
    char *match = 0;    // initialize to null

    // start some loop
    for (int i=0;i<3;i++) {       

        // fetch the next best value...
        int returnValue = bar(&best);       
        // some check (if best is really best!)
        if (returnValue > 10) {
            // if match has previously been populated, free it
            if(match) {
                free(match);
            }
            // save the new best value
            match = best;   
        }
        else {
           // not saving best in match, so free it!
           free(best);
        }
    }
    // end some loop

    // only do this if match was successful?!?
    if(match) {
        printf("%s", match);    
        // clean up once the best of the best has been used...
        free(match);
    }
    return 0; 
}