What is the output of this code ?

Question

plz explain the output of this code thanks in advance
https://code.sololearn.com/cbly2U5snaHT/?ref=app

Answer 1

I have seen this que in our challenges 😊😊😊...
there r two types of que ....
1st one is the one u have coded in which fun is assigned value 30 and then called by & operator and returns value 10....
so output is 10..
2nd one is like this :-
int &fun(){
static int x=10;
return x;
}
int main(){
fun()=30;
cout<<fun();
return 0;
}
outputs:---30
The difference between them is just that in 2nd one the x is made static...
if we don't use static, var. is reinitialized....
u can understand rest.... 😊😊😊
I hope this helps

Answer 2

#include <iostream>
using namespace std;
int & fun();
int main() {
fun()=30; // Assigns 30 to the deleted variable. You have no guarantee it will stay 30 until you use it again. Setting it to 30 could also corrupt a variable of another process.
cout<<fun(); // Gets again a deleted 10 (again a disaster) and puts it into the output stream. No guarantee it will be 10 when the library processes it to write it to the screen.
return 0; //unnecessary
}
int & fun(){ // Returns x by referene. This is a disaster: When x goes out of scope in line 12, it is deleted. But you're still using it.
int x=10;
return x;
}

Answer 3

while calling func() function with reference. the value of &func() becomes 30.and we have declared static keyword with the variable x assigned to 10.then we are returing x.so value of x becomes 30.then the cout statement is displayed.

Answer 4

With a static variable, there is also no disaster. ( :
But the output of version 1 is just probably 10 - it's undefined.

Answer 5

ans is 17