What would happen if copy constructor collects the argument with non-reference type?

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#include "stdafx.h"
#include "iostream"
using namespace std;
class A
{
	int x;
	int y;
public:
	A()
	{
		cout << "Inside default constructor\n" << endl;
		x = 99;
		y = 100;
	}
	void show_data();
	A(A&);
};
A::A(A &temp)
{
	cout << "Inside Copy constructor" << endl;
	x = temp.x;
	y = temp.y;
}
void A::show_data()
{
	cout << x << ' ' << y << endl;
}
int main()
{
	class A a1, b1(a1);
	b1.show_data();
    return 0;
}

Below code throws error:-

// ConsoleApplication81.cpp : Defines the entry point for the console application.
//

#include "stdafx.h"
#include "iostream"
using namespace std;
class A
{
	int x;
	int y;
public:
	A()
	{
		cout << "Inside default constructor\n" << endl;
		x = 99;
		y = 100;
	}
	void show_data();
	A(A);
};
A::A(A temp)
{
	cout << "Inside Copy constructor" << endl;
	x = temp.x;
	y = temp.y;
}
void A::show_data()
{
	cout << x << ' ' << y << endl;
}
int main()
{
	class A a1, b1(a1);
	b1.show_data();
    return 0;
}

error that i am getting is:-

copy constructor for class "A" may not have a parameter of type "A"

What I have tried:

I have created my own copy constructor like below using argument as reference type and it works fine.but if i am trying with non-reference type then it is throwing an error.Just trying to understand why is it so.

Posted 26-Jul-21 21:20pm

Shujaul Hind

Updated 26-Jul-21 22:39pm

v3

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Comments

OriginalGriff 27-Jul-21 3:36am

What error?
Where?
What have you tried to fix it?

Shujaul Hind 27-Jul-21 4:28am

question updated please check

2 solutions

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Richard MacCutchan · Accepted Answer · 2021-07-26T22:16:00

Solution 1

Because the rules of the language state that the copy constructor requires a reference item: Copy constructors - cppreference.com[^].

Posted 26-Jul-21 22:16pm

Richard MacCutchan

Comments

Shujaul Hind 27-Jul-21 4:25am

@Richard you are right but i want to know the reason behind this rule

Richard MacCutchan 27-Jul-21 4:37am

Because without the reference operator the code will pass a copy of the complete object (like a value) in the call, not just a pointer to it.

11917640 Member · Accepted Answer · 2021-07-26T22:39:00

Hypothetical copy constructor with non-reference type, such as

A(A);

requires the copy of its argument. Well, how class instances are passed by value in C++? Using copy constructor. What copy constructor? Obviously, not the same one, this creates an endless recursion.

Standard requires copy constructor with const reference, which gives the desired effect (parameter is read-only), gives a good performance and doesn't contain logical problems as mentioned above.