How does the this c++ code work -

here code

int& fun(){     static int x = 10;     return x; }  int main() {      fun() = 30;      cout<< fun();      getch(); } 

the output 30. how work?

let's read line line:

int& fun() 

declares function named fun returns reference integer variable

{     static int x = 10; 

the x variable static inside function. special place in memory reserved , initialized 10. every time function called x value stored in special location.

    return x; } 

we return x , leave function. let's go main one:

int main() {     fun() = 30; 

remember fun returns reference int. here modify integer 30. since integer has static allocation every time fun called on x start 30, unless other changes made.

    cout<< fun(); 

since x 30 that's get.

    getch(); } 

suggestion: use gdb , trace program's execution step-by step:

$ gdb ./a.out (gdb) b main breakpoint 1 @ 0x40077b: file 1.cpp, line 11. (gdb) r starting program: /tmp/a.out   breakpoint 1, main () @ 1.cpp:11 11      fun() = 30; 

we start gdb, set breakpoint @ begining of main , start program.

(gdb) disp fun() 1: fun() = (int &) @0x60104c: 10 

since fun returns reference static variable can display it's value @ each step in gdb.

(gdb) s fun () @ 1.cpp:6 6       return x; 1: fun() = (int &) @0x60104c: 10 

running single step see in func. place x returned (as reference) attributed 30.

(gdb) n 7   } 1: fun() = (int &) @0x60104c: 30 

indeed, after leaving function, x 30.

(gdb)  main () @ 1.cpp:13 13      cout<< fun(); 1: fun() = (int &) @0x60104c: 30 (gdb) s fun () @ 1.cpp:6 6       return x; 1: fun() = (int &) @0x60104c: 30 (gdb)  7   } 1: fun() = (int &) @0x60104c: 30 (gdb)  main () @ 1.cpp:15 15  } 1: fun() = (int &) @0x60104c: 30 (gdb) q 

we continue program's execution , leave gdb (though question answered).