Predict the output of the below code snippet.

#include <iostream>
using namespace std;

int i;

class A
{
public:
~A()
{
i=21;
}
};

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 21?
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 21, the current value of i gets copied & hence the output is i = 3.

How to make the program to output “i = 21” ?
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 = 21”.

#include <iostream>
using namespace std;

int i;

class A
{
public:
~A()
{
i = 21;
}
};

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 = 21”.

2. Create the object ob in a block scope

#include <iostream>
using namespace std;

int i;

class A
{
public:
~A()
{
i = 21;
}
};

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 21. Finally 21 will copied to the return value.