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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,

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1
  • Technically, what you have are not "multi-dimensional arrays". You have an array of pointers, each of which points to the beginning of a smaller-dimension array. "Multi-dimensional array" refers to something different, with all the memory allocated contiguously. Commented May 1, 2009 at 4:15

5 Answers 5

Reset to default
6

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.

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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...

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8 Comments

In standard C I don't think you can declare 'int i' after the call to make(&array) (or again after the malloc in the make() function).
I compiled it using default gcc options. I thinks this was added to the standard long time ago... But I can't decalare int inside the loop without -C99...
Even compiling with: gcc arr.c -Wall -strict -ansi or gcc arr.c -std=c88 works without any warnings or errors
gcc arr.c -std=c89 of course ;) -std=c88 wont run the compiler ;P
Remember to free the array, too, not just the strings in the array.
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4

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.
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Comments

3

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 = ...

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0

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

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