Method that returns uint16_t array properly in C++

Question

I have my C# code that returns uint array but I want to do it in C++. I looked other posts; they use uint pointer array where my array is not. Does anyone know how to return uint16_t array properly?

This is C# code works fine

  public static UInt16[] GetIntArrayFromByteArray(byte[] byteArray)
        {
            if ((byteArray.Length % 2) == 1)
                Array.Resize(ref byteArray, byteArray.Length + 1);


            UInt16[] intArray = new UInt16[byteArray.Length / 2];


            for (int i = 0; i < byteArray.Length; i += 2)
                intArray[i / 2] = (UInt16)((byteArray[i] << 8) | byteArray[i + 1]);


            return intArray;
        }

This is C++ code that creates syntax error

uint16_t[] GetIntArrayFromByteArray(byte[] byteArray)
{
    //if ((byteArray.Length % 2) == 1)
        //Array.Resize(ref byteArray, byteArray.Length + 1);


    uint16_t[] intArray = new uint16_t[10];


    for (int i = 0; i < 10; i += 2)
        intArray[i / 2] = (uint16_t)((byteArray[i] << 8) | byteArray[i + 1]);


    return intArray;
}

Answer 1

Do not use Type[] ever. Use std::vector:

std::vector<uint16_t> GetIntArrayFromByteArray(std::vector<byte> byteArray)
{
    // If the number of bytes is not even, put a zero at the end
    if ((byteArray.size() % 2) == 1)
        byteArray.push_back(0);


    std::vector<uint16_t> intArray;

    for (int i = 0; i < byteArray.size(); i += 2)
        intArray.push_back((uint16_t)((byteArray[i] << 8) | byteArray[i + 1]));

    return intArray;
}

You can also use std::array<Type, Size> if the array would be fixed size.

More optimal version (thanks to @Aconcagua) (demo)

Here is a full code with more optimal version that doesn't copy or alter the input. This is better if you'll have long input arrays. It's possible to write it shorter, but I wanted to keep it verbose and beginner-friendly.

#include <iostream>
#include <vector>

using byte = unsigned char;

std::vector<uint16_t> GetIntArrayFromByteArray(const std::vector<byte>& byteArray)
{
    const int inputSize = byteArray.size();
    const bool inputIsOddCount = inputSize % 2 != 0;
    const int finalSize = (int)(inputSize/2.0 + 0.5);
    // Ignore the last odd item in loop and handle it later
    const int loopLength = inputIsOddCount ? inputSize - 1 : inputSize;

    std::vector<uint16_t> intArray;
    // Reserve space for all items
    intArray.reserve(finalSize);
    for (int i = 0; i < loopLength; i += 2) 
    {
      intArray.push_back((uint16_t)((byteArray[i] << 8) | byteArray[i + 1]));
    }

    // If the input was odd-count, we still have one byte to add, along with a zero
    if(inputIsOddCount) 
    {
      // The zero in this expression is redundant but illustrative
      intArray.push_back((uint16_t)((byteArray[inputSize-1] << 8) | 0));
    }
    return intArray;
}

int main() {
    const std::vector<byte> numbers{2,0,0,0,1,0,0,1};
    const std::vector<uint16_t> result(GetIntArrayFromByteArray(numbers));

    for(uint16_t num: result) {
        std::cout << num << "\n";
    }

    return 0;
}