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Home/ Questions/Q 6800173
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T18:56:14+00:00 2026-05-26T18:56:14+00:00

This seems like it should have a super easy solution, but I just can’t

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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.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T18:56:14+00:00Added an answer on May 26, 2026 at 6:56 pm

    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.

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