Copying from One Dynamically Allocated Array to Another C++

Question

This seems like it should have a super easy solution, but I just can't figure it out. I am simply creating a resized array and trying to copy all the original values over, and then finally deleting the old array to free the memory.

void ResizeArray(int *orig, int size) {
    int *resized = new int[size * 2]; 
    for (int i = 0; i < size; i ++)
        resized[i] = orig[i];
    delete [] orig;
    orig = resized;
}

What seems to be happening here is that resized[i] = orig[i] is copying values by reference rather than value, as printing orig after it gets resized returns a bunch of junk values unless I comment out delete [] orig. How can I make a deep copy from orig to resized, or is there some other problem that I am facing? I do not want to use std::vector.

Solution 1

Remember, parameters in C++ are passed by value. You are assigning resized to a copy of the pointer that was passed to you, the pointer outside the function remains the same.

You should either use a double indirection (or a "double pointer", i.e. a pointer to a pointer to int):

void ResizeArray(int **orig, int size) {
    int *resized = new int[size * 2]; 
    for (int i = 0; i < size; i ++)
        resized[i] = (*orig)[i];
    delete [] *orig;
    *orig = resized;
}

or a reference to the pointer:

void ResizeArray(int *&orig, int size) {
    int *resized = new int[size * 2]; 
    for (int i = 0; i < size; i ++)
        resized[i] = orig[i];
    delete [] orig;
    orig = resized;
}

By the way, for array sizes you should use the type std::size_t from <cstddef> - it is guaranteed to hold the size for any object and makes clear that we are dealing with the size of an object.

Solution 2

I highly suggest replacing the arrays with std::vector<int>. This data structure will resize as needed and the resizing has already been tested.

Solution 3

orig must be a pointer to a pointer to assign it to resized:

int **orig;
*orig = resized;

Comments

[André Caron]

+1 The other option is to return the new array using... the return value.

[Matteo Italia]

@AndréCaron: that's an option too, but I think that these forms suit better the indented use.

[Michael Price]

As a point of clarification, by "double pointer" the respondent means "double indirection". When I first read it, I was confused why he wanted to substitute double * for int *.

[Markus]

While vectors in general are something interesting, there also exist a variety of reason against them. Thus, your answer is opinion-based and rather a comment than an actual answer to OP's problem.

[Thomas Matthews]

@Markus: One could say the same about arrays. There also exists a variety of reasons against them, thus the OP's issue is opinion based and should be closed (and maybe deleted).