How to free an array of char-pointer?

Question

I use this Method to convert values from a list into an array for use in an execvp()-Systemcall:

char **list2argarray(struct shellvalue *values, int count)
{
    char **array = (char **)malloc((count + 1) * sizeof(char *));
    int i = 0;

    while (values)
    {
        char *word = values->word;

        array[i] = (char *)malloc(sizeof(word) + 1);
        strcpy(array[i], word);
        values = values->next;
        i++;
    }
    array[i] = NULL;
    return array;
}

What is a proper way to free such Arrays? I tried it with things like

void freeargpointer(char **array, int count)
{
    int i = 0;

    while (*array)
    {
        free(*array);
        (*array)++;
    }
}

But everytime when i reach the free-syscall, while debugging, the programm crashes with errors like this one:

free(): invalid next size (fast): 0x000000000060c180 ****

Solution 1

The problem is that (*array)++ doesn't give you the next pointer you allocated, so you can't free it. Your free routine should be:

void freeargpointer(char** array)
{
    int i;

    for ( i = 0; array[i]; i++ )
        free( array[i] );

    free( array );
}

Or, similarly,

void freeargpointer(char** array)
{
    char **a;

    for ( a = array; *a; a++ )
        free( *a );

    free( array );
}

NOTE: I removed the count argument since it is unnecessary.

Solution 2

This line is wrong.

(*array)++;

You need to have.

++array;

Comments

[Wai Ha Lee]

Where is count used?

[MattSom]

So where is count used?

[lurker]

@MattSom it's not. I carried it along unwittingly from the OP's original code. I'll remove it for clarity.

[chux - Reinstate Monica]

Could add about incorrect array[i] = (char *)malloc(sizeof(word) + 1);. Likely needs array[i] = (char *)malloc(strlen(word) + 1);