Array initialization functions

Question

I was playing around with C++ and I stumbled upon this problem. I'm trying to initialize an array pointer on the heap, and it works inside the initialize(), where it outputs 69, but in the main(), it crashes with the error EXC_BAD_ACCESS.

#include <iostream>

void initialize(int* array, int size) {
    array = new int[size];

    // Testing
    array[2] = 69;
    std::cout << array[2] << std::endl; // Works fine
}

int main() {

    int size = 3;
    int* array;

    // Initializing
    initialize(array, size);

    // Testing
    std::cout << array[2] << std::endl; // Crash, EXC_BAD_ACCESS

    // Cleanup
    delete[] array;
    array = nullptr;


    return EXIT_SUCCESS;
}

Please help me understand the problem with this.

Yes, I know I should use std::vector but I want to understand why this doesn't work :)

For that matter, it crashes at the delete[] function too, with the error text "pointer being freed was not allocated" — Mads Marquart
– Mads Marquart, Commented Oct 20, 2015 at 17:51
You pass your array pointer to your function by value - so when you assign new int[size] to it, array in main stays unchanged. — melak47
– melak47, Commented Oct 20, 2015 at 17:52
@MadsMarquart Well, I'm pretty sure we have a duplicate for this question, I see it asked about 3-5 times a month. Though there might be too many duplicates for it, and we don't have a real canonical Q&A. — πάντα ῥεῖ
– πάντα ῥεῖ, Commented Oct 20, 2015 at 18:07

cadaniluk · Accepted Answer · 2015-12-14 14:23:04Z

18

When you pass array to the function, a copy of that pointer is made. When you assign new int[size]; to array, you assign it actually to the argument, which is the copy I was talking about. To really modify the array defined in main, use references. Change the definition of the function to

void initialize(int*& array, int size)

or return the pointer like¹

int* initialize(int size)

and try it again.

I recommend the second method due to its higher expressiveness: something like

initialize(array, 3);

does not make clear if array is modified or not. OTOH,

int* array = initialize(3);

does.

_{¹ as noted by @Jack in the comments to this answer}

edited Dec 14, 2015 at 14:23

answered Oct 20, 2015 at 17:52

cadaniluk

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3 Comments

Jack Over a year ago

Or directly change the signature to int* initialize(int size) if you don't have any specific purpose to set an existing pointer by reference.

gmoniava Over a year ago

"array to the function, a copy of it it is made."-you may want to clear out that copy of the pointer is made not the whole array itself

cadaniluk Over a year ago

@Giorgi Done, although I think there wasn't much to confuse.

dspfnder · Accepted Answer · 2015-10-20 18:10:25Z

0

The reason why the program fails is because you want the memory to be allocated outside the initialize function and the function to operate on that memory.

Simply remove the new statement from your function so that it looks like this...

void initialize(int* array, int size) {
        for (int i = 0; i < size; i++) {
        cout << array[i] << " ";
    }
}

... then, do your allocation in main and just before the function call...

int size = 3;
int* array = new int [size];

initialize(array, size);

answered Oct 20, 2015 at 18:10

dspfnder

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Comments

Pravin Battu · Accepted Answer · 2015-10-22 17:32:36Z

0

Pass the address of the pointer to avoid the error message

answered Oct 22, 2015 at 17:32

Pravin Battu

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