Vfrdtyky

This is simple program that is supposed to handle a dynamic array of numbers and then to filter out the elements that are even and put them into a new array and then print both arrays on the screen.

This is the header file:

#pragma once





namespace filter

{

    class Array

    {

        double *arr;

        int n;



public:

    Array();

    Array(int);

    Array(const Array&);

    Array(Array&&);

    void PutIn();

    void PrintOut() const;

    Array isEven();

    Array filter(const std::function<bool(int)>&) const;

    ~Array();



};

}

Then, this is the implementation of functions:

#include <iostream>

#include <functional>

#include "Array.h";

using namespace filter;

using namespace std;



Array::Array() :arr(nullptr), n(0)

{ }



Array::Array(int n)

{

    this->n = n;

    arr = new double[n];

}



Array::Array(const Array &a1)

{

    n = a1.n;

    arr = new double[n];

    for (int i = 0; i < n; i++)

        arr[i] = a1.arr[i];

}



Array::Array(Array &&a1)

{

    n = a1.n;

    for (int i = 0; i < n; i++)

        arr[i] = a1.arr[i];

    a1.n = 0;

    a1.arr = nullptr;

}



void Array::PutIn()

{

    cout << "Insert elements:n";

        for (int i = 0; i < n; i++)

            cin >> arr[i];



}

void Array::PrintOut() const

{

    cout << "nYour array is :n";

    for (int i = 0; i < n; i++)

        cout << arr[i] << "t";

}



Array Array::isEven()

{

    return filter((int x) { return x % 2; });

}

Array Array::filter(const std::function<bool(int)> &f) const

{

    int b = 0;



    for (int i = 0; i < n; i++)

        if (f(arr[i]) == 0)

            b++;



    Array temp(b);

    b = 0;

    for (int i = 0; i < n; i++)

        if (f(arr[i]) == 0)

        {

            temp.arr[b] = arr[i];

            b++;

        }

    return temp;

}



Array::~Array()

{

    deletearr;

    n = 0;

}

Finally, this is the source code:

#include <iostream>

#include <functional>

#include "Array.h"

using namespace filter;

using namespace std;









int main()

{



    Array a(5);



    a.PutIn();



    Array b = a.isEven();    //WHY THIS LINE OF CODE INVOKES MOVE CONSTRUCTOR AND NOT COPY CONSTRUCTOR?



    a.PrintOut();

    b.PrintOut();



    getchar();

    getchar();



}

So, as you can see, this is relatively simple program that needs to handle an array with five elements in it entered by user and then to create a new array that consists of even elements of the first array. When i run this, it works fine, however, there is one little thing that i don't understand here.

If you look at the source code, notice the line where i left my comment, that is the line where move constructor is called, but i don't know why. That would mean that a.IsEven() is a RVALUE, since move constructor works with RVALUES, right? Can anyone explain me why this is rvalue and what is the correct way to understand this? Any help appreciated!

Your assumption that calling isEven invokes your move constructor is not in fact correct. Nor does it invoke your copy constructor. Instead, RVO ensures that the object returned is constructed directly at the calling site so neither is required.

Live Demo (which doesn't address any of the flaws in your code mentioned in the comments).