Creating dynamically allocated arrays using Xcode's g++

Question

So, I recently tried doing this:

void Foo(int **bar, int size) {
    bar = new int *[size];
    for (int i = 0; i < size; i++)
        bar[i] = new int[4];
    // Do other stuff
    return;
}

int main(int argc, char *argv[]) {
    int ** data;
    int size = atoi(argv[1]);
    Foo(data, size);
    for (int i = 0; i < size; i++) {
        for (int j = 0; j < 4; j++)
            cout << "i = " << i << ", j = " << j << ", data = " << data[i][j] << endl;
        // Do other stuff
    }

    for (int i = 0; i < size; i++)
        delete[] data[i];
    delete[] data;
    return 0;
}

and invariably I would get a segfault right at that cout statement. After changing the code so that the array was dynamically allocated in main the problem went away. So, the difference between a segfault is whether or not an array is dynamically allocated and destroyed in the same function. This seems wrong to me as this shouldn't be a problem with traditional C++. I am using a MacBook with Xcode for the g++ command. Can anyone else confirm that Xcode's implementation does this?

Your code invokes undefined behavior due to pointer mismanagement. There is nothing wrong with your C++ compiler or installation. — PaulMcKenzie
– PaulMcKenzie, Commented Nov 15, 2015 at 6:28
If you had inspected the value of data after the call to Foo, you should have seen that data didn't change at all. That's what all the answers given to you will explain. Also: stackoverflow.com/questions/27487495/… — PaulMcKenzie
– PaulMcKenzie, Commented Nov 15, 2015 at 6:35
@PaulMcKenzie I did inspect data after I called Foo and I did notice that it hadn't changed. However, I was under the impression that sending any pointer to a function was a pass-by-reference. And the answer given below does explain my problem, thank you — Woody1193
– Woody1193, Commented Nov 15, 2015 at 15:58
A better example is if you set data to NULL before you called Foo. You would have seen that data would still be NULL, thus your cout would be working with a NULL pointer (undefined behavior). And no, sending a pointer is pass-by-value, no different than any other value type being passed. Pass-by-reference in C++ means exactly that -- passing a reference. A pointer is not a reference. — PaulMcKenzie
– PaulMcKenzie, Commented Nov 15, 2015 at 18:10

Paul R · Accepted Answer · 2015-11-15 06:32:57Z

1

It's nothing specific to Xcode or g++ - just a bug in your code - you need to pass bar by reference in order to use it as an output parameter.

Change:

void Foo(int **bar, int size)

to:

void Foo(int ** &bar, int size)

Note that in general it's preferable to use e.g. std::vector rather than naked C-style pointers for arrays etc.

edited Nov 15, 2015 at 6:32

answered Nov 15, 2015 at 6:27

Paul R

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1 Comment

Woody1193 Over a year ago

I honestly didn't know you could do that

Eugene · Accepted Answer · 2015-11-15 06:28:41Z

0

You pass data to Foo by value, so data in main () remains uninitialized when you allocate memory inside Foo.

answered Nov 15, 2015 at 6:28

Eugene

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