can anyone explain the output?

Question

Can anyone explain the output?

#include<iostream>

using namespace std;  

int &fun(){   
  static int x = 10;   
  return x;   
} 

int main(){         
  fun() = 30;
  cout << fun();          
  return 0;         
}

output is 30

Answer 1

That's how static locals work - they persist the value between the function calls. Basically fun() has a static local and returns a reference to it, the effect is roughly the same as you would have with a global variable.

Answer 2

You return the static by reference, so when you do fun() = 30 you change it.

It's pretty clear, no?

Basically, foo() returns a reference to x .

Answer 3

When you call fun() a static variable is created and you return the reference to it. Basically, because of the static , the variable is not destroyed even if you exit the scope of the function. You affect the reference with 30 and then recalling the function you get 30 (the x at the second call is exactly the same at the first call). Basically the static works like a global variable in this case.

Answer 4

AS fun is the reference to the function so when you write this line fun() = 30; it stores 30 in its return value ie x , that is why you are getting the output as 30.