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!

Kaidjin · Accepted Answer · 2014-03-12 17:06:45Z

7

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.

answered Mar 12, 2014 at 17:06

Kaidjin

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1 Comment

Joachim Isaksson Over a year ago

+1 Calling free on the constants is probably not a great idea either, which this fixes too.

Paul R · Accepted Answer · 2014-03-12 17:05:37Z

6

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");

answered Mar 12, 2014 at 17:05

Paul R

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