Correct free() of string array C

Question

This may be a newbie fault, but I am greatful for any tips on what exactly I do wrong.

The code:

int main()
{
  int i =0;
  char ** nameslist;
  nameslist = malloc(5 * sizeof(char*));

  for (i=0; i <5; i++)
  {
      nameslist[i]=malloc((20+1)*sizeof(char));
  }

  nameslist[0]="John";
  nameslist[1]="Adam";
  nameslist[2]="Nick";
  nameslist[3]="Joe";
  nameslist[4]="Peter";

  for (i=0; i <5; i++)
  {
    free(nameslist[i]);
  }

  free(nameslist);
  return 0;
}

Using Valgrind, I still see that I have memory leaks in heap - how do I fix this issue? I suspect that I malloc too much space - but still, how do I go about freeing space I don't necesarrily use?

Thankful for any hints!

Answer 1

The problem is that when you write this:

nameslist[0]="John";

You're not actually using the memory you have allocated. "John" is itself a pointer and you're overwriting the adress returned by malloc with this new pointer.

Use the strncpy() function to copy the string into the memory you allocated.

Answer 2

Your malloc and free stuff is fine, but you are orphaning the original string when you try to do assignment like this:

nameslist[0]="John";

Instead you should use strcpy:

strcpy(nameslist[0], "John");