Playing with Destructors in C++

Predict the output of the below code snippet. CPP

#include <iostream>
using namespace std;

int i;

class A
{
public:
    ~A()
    {
        i=10;
    }
};

int foo()
{
    i=3;
    A ob;
    return i;
}

int main()
{
    cout << "i = " << foo() << endl;
    return 0;
}

Output of the above program is "i = 3". Why the output is i= 3 and not 10? While returning from a function, destructor is the last method to be executed. The destructor for the object "ob" is called after the value of i is copied to the return value of the function. So, before destructor could change the value of i to 10, the current value of i gets copied & hence the output is i = 3. How to make the program to output "i = 10" ? Following are two ways of returning updated value: 1) Return by Reference: Since reference gives the l-value of the variable,by using return by reference the program will output "i = 10". CPP

#include <iostream>
using namespace std;

int i;

class A
{
public:
	~A()
	{
		i = 10;
	}
};

int& foo()
{
	i = 3;
	A ob;
	return i;
}

int main()
{
	cout << "i = " << foo() << endl;
	return 0;
}

The function foo() returns the l-value of the variable i. So, the address of i will be copied in the return value. Since, the references are automatically dereferened. It will output "i = 10".
2. Create the object ob in a block scope CPP

#include <iostream>
using namespace std;

int i;

class A
{
public:
    ~A()
    {
        i = 10;
    }
};

int foo()
{
    i = 3;
    {
        A ob;
    }
    return i;
}

int main()
{
    cout << "i = " << foo() << endl;
    return 0;
}

Since the object ob is created in the block scope, the destructor of the object will be called after the block ends, thereby changing the value of i to 10. Finally 10 will copied to the return value.
This article is compiled by Aashish Barnwal and reviewed by GeeksforGeeks team.