Accessing variables with the same name at different scopes [duplicate]

问题

With

#include <iostream>
using namespace std;

        int a = 1;

int main()
{
        int a = 2;

        if(true)
        {
                int a = 3;
                cout << a 
                     << " " << ::a // Can I access a = 2 here?
                     << " " << ::a << endl;
        }
        cout << a << " " << ::a << endl;
}

having the output

3 1 1
2 1

Is there a way to access the 'a' equal to 2 inside the if statement where there is the 'a' equal to 3, with the output

3 2 1
2 1

Note: I know this should not be done (and the code should not get to the point where I need to ask). This question is more "can it be done".

回答1:

No you can't, a (2) is hidden.

Ref: 3.3.7/1

A name can be hidden by an explicit declaration of that same name in a nested declarative region or derived class (10.2).

Ref: 3.4.3/1

The name of a class or namespace member can be referred to after the :: scope resolution operator (5.1) applied to a nested-name-specifier that nominates its class or namespace. During the lookup for a name preceding the :: scope resolution operator, object, function, and enumerator names are ignored. If the name found is not a class-name (clause 9) or namespace-name (7.3.1), the program is ill-formed.

回答2:

Short answer: no. You are basically overriding the inherited scope of a locally, and it will use that local copy over any inherited ones.

Basically like a child object overriding a function or variable of the parent object, it will use it's copy of it regardless of what the parent had.

来源：https://stackoverflow.com/questions/5187891/accessing-variables-with-the-same-name-at-different-scopes

标签

c++

scope