In C++, what is the scope resolution ("order of precedence") for shadowed variable names?

Question

In C++, what is the scope resolution ("order of precedence") for shadowed variable names? I can't seem to find a concise answer online.

For example:

#include <iostream>

int shadowed = 1;

struct Foo
{
    Foo() : shadowed(2) {}

    void bar(int shadowed = 3)
    {
        std::cout << shadowed << std::endl;
            // What does this output?

        {
            int shadowed = 4;
            std::cout << shadowed << std::endl;
                // What does this output?
        }
    }

    int shadowed;
};


int main()
{
    Foo().bar();
}

I can't think of any other scopes where a variable might conflict. Please let me know if I missed one.

What is the order of priority for all four shadow variables when inside the bar member function?

Answer 1

Your first example outputs 3. Your second outputs 4.

The general rule of thumb is that lookup proceeds from the "most local" to the "least local" variable. Therefore, precedence here is block -> local -> class -> global.

You can also access each most versions of the shadowed variable explicitly:

// See http://ideone.com/p8Ud5n
#include <iostream>

int shadowed = 1;

struct Foo
{
    int shadowed;
    Foo() : shadowed(2) {}
    void bar(int shadowed = 3);
};

void Foo::bar(int shadowed)
{
    std::cout << ::shadowed << std::endl; //Prints 1
    std::cout << this->shadowed << std::endl; //Prints 2
    std::cout << shadowed << std::endl; //Prints 3
    {
        int shadowed = 4;
        std::cout << ::shadowed << std::endl; //Prints 1
        std::cout << this->shadowed << std::endl; //Prints 2
        //It is not possible to print the argument version of shadowed
        //here.
        std::cout << shadowed << std::endl; //Prints 4
    }
}

int main()
{
    Foo().bar();
}