How to free() the memory which is referenced by **pointer?

Question

Please help me understand this:

I have a pointer to a pointer
```
char** word_array = NULL;
```

Then I dynamically allocate memory in my code:

n = 6;
word_array = (char *) malloc(n* sizeof(char*));

How do I delete all the memory allocated for the "array" of pointers? Is this call to free right?
```
free(word_array);
```
Or should I make a loop:
```
for(int i = 0 ; i < n; i++) free(word_array + i);
```

Take a glance, for it may help: https://stackoverflow.com/a/53659983/3808573 — ssd, Oct 04 '20 at 13:35

score 4 · Accepted Answer · edited Oct 04 '20 at 13:26

4

You used one malloc, so you should use one free.

Also the cast to char* is nonsense: word_array has type char**, not char*, and casting result of malloc() is discouraged.

So the entire flow will be like this:

int n;

char** word_array = NULL;

n = 6;
word_array = malloc(n* sizeof(char*));

if (word_array == NULL) {
    /* handle allocation error */
} else {
    /* do some work with word_array */

    /* free pointers stored in word_array if they are dynamically allocated
       and not freed yet */

    free(word_array);
}

edited Oct 04 '20 at 13:26

klutt

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answered Oct 04 '20 at 13:17

MikeCAT

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2

And even better: `sizeof(*word_array)` – klutt Oct 04 '20 at 13:19
And *even better*: don’t declare `word_array` (or `n`) before assigning to it. – Konrad Rudolph Oct 04 '20 at 13:20
@MikeCAT Thanks for the reply! yes, I wasn't intending to add the casting, Thanks for mentioning it as well! – PETROSCHOOL Oct 04 '20 at 13:23
@PETROSCHOOL Sorry, but I really have to ask you you did a cast without intention? :D – klutt Oct 04 '20 at 13:25
@klutt yes, and it's unnecessary here – PETROSCHOOL Oct 04 '20 at 13:46
@PETROSCHOOL Yes, I'm just asking how you managed to add it by accident. – klutt Oct 04 '20 at 14:40

How to free() the memory which is referenced by **pointer?

1 Answers1