C memory management error?

Question

This is my C program: ``` #include #include #include #include #include #define FALSE 0 #define TRUE 1 typedef struct _Frag { struct _Frag *next; char *seq; int x1; int length; } Frag; typedef struct _Fragment { int type; Frag *frag_list; } Fragment; static void free_frags (Fragment * frags, int len) { int i; for (i = 0; i < len; i++) { Fragment *fragment = &frags[i]; Frag *current = fragment->frag_list; while (current != NULL) { free (current->seq); fragment->frag_list = current->next; free (current); current = fragment->frag_list; } /* to do : free fragment */ free (fragment); fragment = NULL; } free (frags); } int main () { Fragment *frags = (Fragment *) malloc (10 * sizeof (Fragment)); int i, j; for (i = 0; i < 10; i++) { Fragment *fragment = &frags[i]; fragment->frag_list = (Frag *) malloc (1 * sizeof (Frag)); Frag *frag = fragment->frag_list; frag->seq = malloc (6 * sizeof (char)); strcpy (frag->seq, "55555"); frag->next = (Frag…

Accepted Answer

You have multiple problems.

Here, you allocate space for 5 chars but copy 6 in (the nul-terminator at the end of the string needs a space, too):

```
  frag->seq = malloc (5 * sizeof (char));
  strcpy (frag->seq, "55555");
```

At the same point, you never set `frag->next` in the second `frag` you allocate. You need to set it to `NULL`, so that the `while` loop in the `free_frag` routine doesn't run off into the weeds.

The third problem is here:

```
  /* to do : free fragment */
  free (fragment);
```

You free `fragment`, but it's not a whole block recieved from `malloc` - it's just one of the single block of 10 fragments you allocated in one go. The later `free(frags)` frees that block properly, so you just need to remove that buggy line.

C memory management error?

Solution

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