Output with reason

Question

#include <iostream>
using namespace std;
int func(int) ;
int num = 10 ;

int main()
{
int num ;
num = 5 ;
cout << num ;

cout << func(num) ;
}

int func(int x)
{
return num ;
}

Answer 1

# include<...>
using........

global function declared int func(int)
global var declared int num = 10

main func()
{
...
another num variable declared..but only main can use it
}// here main ends...
outside main block..num means that top global var that is equal to10

func()
{
.....here returning num means returnig that global var.
global var can be accessed to any block of function if there is no other same var name declared..

}

Answer 2

#include <iostream>
using namespace std;

int func(int) ;

int num = 10 ;
// global variable num

int main()
{
int num ;
num = 5 ;
// local var num for only main() func.
cout << num ;
// prints 5

cout << func(num) ;
//calls func(5) ; argument that is passed is 5..
}

int func(int x)
{
return num ;
// from main thpugh argument that is passed is 5...its copied in x...here x is equal to 5,in this function there is no such declaration for num.
so when it returns num this "num" is that global variable num that is 10..so it prints 10..
}

ans is 510 as output..

✌✌

Answer 3

can you explain it more..sorry i didnt got it

Answer 4

in main...again a same name variable is there...
so it used that local main num...as 5

func has no other num var.
so it uses globally declared num var.and prints 10