How does the this c++ code work [closed] - c++

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Here is the code
int& fun(){
static int x = 10;
return x;
}
int main() {
fun() = 30;
cout<< fun();
getch();
}
The output is 30.
How does this work?

Answer 1

Let's read it line by line:
int& fun()
declares a function named fun which returns a reference to an integer variable
{
static int x = 10;
The x variable is static inside this function. A special place in memory is reserved for it and is initialized with 10. Every time this function is called x will get the value stored in that special location.
return x;
}
We return x and leave the function. Let's go to the main one:
int main()
{
fun() = 30;
Remember that fun returns the reference to an int. Here we modify the integer to be 30. Since this integer has static allocation every time fun will be called from now on x will start with 30, unless other changes are made.
cout<< fun();
Since x is 30 that's what you get.
getch();
}
Suggestion: use GDB and trace the program's execution step-by step:
$ gdb ./a.out
(gdb) b main
Breakpoint 1 at 0x40077b: file 1.cpp, line 11.
(gdb) r
Starting program: /tmp/a.out
Breakpoint 1, main () at 1.cpp:11
11 fun() = 30;
We start GDB, set a breakpoint at the begining of main and start the program.
(gdb) disp fun()
1: fun() = (int &) #0x60104c: 10
Since fun returns a reference to a static variable I can display it's value at each step in GDB.
(gdb) s
fun () at 1.cpp:6
6 return x;
1: fun() = (int &) #0x60104c: 10
Running a single step we see that we are in func. This is the place where x is returned (as a reference) to be attributed 30.
(gdb) n
7 }
1: fun() = (int &) #0x60104c: 30
Indeed, after leaving the function, x is 30.
(gdb)
main () at 1.cpp:13
13 cout<< fun();
1: fun() = (int &) #0x60104c: 30
(gdb) s
fun () at 1.cpp:6
6 return x;
1: fun() = (int &) #0x60104c: 30
(gdb)
7 }
1: fun() = (int &) #0x60104c: 30
(gdb)
main () at 1.cpp:15
15 }
1: fun() = (int &) #0x60104c: 30
(gdb) q
We continue the program's execution and leave GDB (though your question is already answered).

Answer 2

It is straightforward : fun returns the reference to the static variable inside the function, which is assigned to 30 by this line:
fun() = 30;
this means, x inside the function is changed to 30. What is more? It is 30! You print it by calling again fun() in the next line.
Note that a static variable lives till the end of the program: it is not destroyed even if the function returns!

Answer 3

The fun function returns a reference-to an int type. This way, when you call fun() = 30, you actually set x to 30.
This way when you print it out, the fun function will no longer set it, cause that is a one-off declaration.