When I run this code the array has a new size after, is there anything wrong or bad about it ?

static int[] ExpandArray(int[] input, int add_size)
{
    for (int i = 0; i < add_size; i  )
    {
        int[] temp = input;
        input = new int[input.Length   1];
        for (var j = 0; j < temp.Length; j  )
        {
            input[j] = temp[j];
        }
    }
    return input;
}

static void Main(string[] args)
{   
    int[] ovride = new int[3] { 1, 2, 3 };
    ovride = ExpandArray(ovride, 10);
    ovride = ExpandArray(ovride, 10);

    Console.WriteLine(ovride.Length);    
}        

CodePudding user response：

You can use Array.Resize which:

Changes the number of elements of a one-dimensional array to the specified new size.

int[] ovride = new int[3] { 1, 2, 3 };
Array.Resize(ref ovride, ovride.Length   10);
Array.Resize(ref ovride, ovride.Length   10);
Console.WriteLine(ovride.Length); // prints 23

But if you expect collection size changes List can be a more suitable option for your goal.

CodePudding user response：

is there anything wrong or bad about it ?

This isn't code review, but:

1. Yes. You should not resize arrays. This involves a new allocation and a copy of all elements. As does Array.Resize(), by the way.
2. Hey, there is a method that already does this: Array.Resize(). Don't reinvent the wheel.
3. You definitely should not do the resize in a loop.

So to clean up the code a little:

static int[] ExpandArray(int[] input, int sizeToAdd)
{
    // Create a new array with the desired size
    var ouput = new int[input.Length   sizeToAdd];
    
    // Copy all elements from input to output
    for (int i = 0; i < input.Length; i  )
    {
        output[i] = input[i];
    }

    // Return the new array, having the remaining 
    // items set to their default (0 for int)
    return output;
}

You'd actually want input to be updatable by ref, and then end with input = output.

Ultimately, just use a List<int>, as that allows for more efficient resizing, and does so automatically when necessary.