Dynamic allocation in C - Allocate into a function

Question

What I want exactly is to allocate tab and fill it with some numbers.

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

void prepare_internship(int** tab) {
    *tab = malloc(5 * sizeof(int));
    *(*tab+0) = 1;
}

int main() {
    int** tab;
    prepare_internship(tab);
    printf("%d", *(tab));
    return 0;
}

[Please see this discussion on why not to cast the return value of `malloc()` and family in `C`.](http://stackoverflow.com/q/605845/2173917). — Sourav Ghosh, Dec 22 '15 at 14:05
At which line does it crash when you step through the program in your debugger? — Lundin, Dec 22 '15 at 14:07
`objects = (object**) malloc(...` : this can't change caller side `objects`. — BLUEPIXY, Dec 22 '15 at 14:10
Ah not that bug again. We had that yesterday too, was also a pointer-to-pointer passed by parameter. Methinks we need a canonical duplicate for this seemingly frequent bug. — Lundin, Dec 22 '15 at 14:17
`int** tab; prepare_internship(tab);` --> `int* tab; prepare_internship(&tab);` — BLUEPIXY, Dec 22 '15 at 14:19
@XPRO An array of pointers is not an array of numbers, nor is it a 2D array of numbers. — Lundin, Dec 22 '15 at 14:22
Ok the question doesn't make any sense any longer after the radical edit... — Lundin, Dec 22 '15 at 14:25

interjay · Answer 1 · 2015-12-22T14:24:02.743

3

You're passing an uninitialized pointer into prepare_internship. What you need instead if you want the function to modify tab is:

int* tab;
prepare_internship(&tab);

But in this case it's better to change your function to have a int* return type and then you can simply call it with tab = prepare_internship();

edited Dec 22 '15 at 14:24

answered Dec 22 '15 at 14:21

interjay

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I need to take speed typing lessons. ;-) – donjuedo Dec 22 '15 at 14:22
hhh Thank you too ☺ but i still don't understand why I should pass an initialized pointer ! – XPRO Dec 22 '15 at 14:26
@XPRO You need to pass in the address of an actual variable that you want the function to change. What your code did was pass in an uninitialized address that points to a random memory position. – interjay Dec 22 '15 at 14:28

score 0 · Answer 2 · answered Dec 22 '15 at 14:21

0

int** tab is not initialized. It should be int* tab and when passed as an argument, passed like this: &tab.

answered Dec 22 '15 at 14:21

donjuedo

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Dynamic allocation in C - Allocate into a function

2 Answers2