Reputation: 500
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 :)
Upvotes: 2
Views: 12962
Reputation: 15229
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
Upvotes: 18
Reputation: 1123
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);
Upvotes: 0