Passing multi-dimensional arrays in C

Question

Hey guys,

I am currently trying to learn C and I have come to a problem that I've been unable to solve.

Consider:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define ELEMENTS 5

void make(char **array, int *array_size) {
    int i;
    char *t = "Hello, World!";

    array = malloc(ELEMENTS * sizeof(char *));

    for (i = 0; i < ELEMENTS; ++i) {
     array[i] = malloc(strlen(t) + 1 * sizeof(char));
     array[i] = strdup(t);
    }
}

int main(int argc, char **argv) {
    char **array;
    int size;
    int i;

    make(array, &size);

    for (i = 0; i < size; ++i) {
     printf("%s\n", array[i]);
    }

    return 0;
}

I have no idea why the above fails to read back the contents of the array after creating it. I have literally spent an hour trying to understand why it fails but have come up empty handed. No doubt it's something trivial.

Cheers,

Answer 1

See PolyThinker's comment which is absolutely spot on.

In addition to the way you pass the array, you should check a few other issues:

1. Perhaps you should assign something to array_size in make(...)?
2. strdup(char*) allocates memory, the malloc for array[i] is not necessary.
3. You should free all the memory you allocate after you don't need it anymore.

Answer 2

You need to pass the address of "array" into the function. That is, you need char ***. This is because you need to change the value of array, by allocating memory to it.

EDIT: Just to make it more complete, in the function declaration you need to have something like

void make(char ***array, int *array_size)

Then you need to call it using

make(&array, &size);

Inside the function make, allocate memory with

*array = malloc(ELEMENTS * sizeof(char *));

And change other places accordingly.

Also, as kauppi has pointed out, strdup will allocate memory for you, so you don't need to do malloc on each string.

Answer 3

size is declared but gets no value assigned (that should happen in fuction make, I suppose).

Answer 4

You are passing the current value of array to make as a copy (on the stack). when you change array in make(), you're only changing the copy, not the actual variable. Try passing by reference with &, or make it a char *** and work with *array = ...

Answer 5

Here is the working code:

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define ELEMENTS 5

void make(char ***array) {
    char *t = "Hello, World!";

    *array = malloc(ELEMENTS * sizeof(char *));

    int i;
    for (i = 0; i < ELEMENTS; ++i) {
        (*array)[i] = strdup(t);
    }
}

int main(int argc, char **argv) {
    char **array;
    make(&array);

    int i;
    for (i = 0; i < ELEMENTS; ++i) {
        printf("%s\n", array[i]);
        free(array[i]);
    }
    free(array);
    return 0;
}

As the other have posted - there was unused size, and strdup allocates memory by itself, and it is nice to free the memory afterwards...