Thread: Void function (What will this program print out?

Shadowing, as in shadowing a parameter?

Code:

void hippo(int var) {
    int var;
}

Code:

#include <iostream>

int x = 20;

int main()
{
    int x = 50;

    std::cout << "Local x: " << x << std::endl << "Global x: " << ::x << std::endl;

    return 0;
}

Output:

Code:

Local x: 50
Global x: 20

Originally Posted by 7stud

Actually, my book says a variable in an inner scope masks one in an outer scope.
It uses hiding in the context of inheritance. I've never heard shadowing before.

Oh god, masking... will the similar terms never end.

In this code fragment the scope operator is used to address a global variable instead of the local variable with the same name. In C++ the scope operator is used extensively, but it is seldomly used to reach a global variable shadowed by an identically named local variable. Its main purpose will be described in chapter 6.

http://www.icce.rug.nl/documents/cpl...lusplus03.html