问题描述

在 C++ 中，shadowed 变量的范围解析(优先顺序")是什么名字?我似乎无法在网上找到简明的答案.

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

例如:

#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();
}

我想不出任何其他变量可能会发生冲突的范围.如果我错过了，请告诉我.

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

bar 成员函数中所有四个 shadow 变量的优先级顺序是什么?

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

推荐答案

您的第一个示例输出 3.您的第二个输出 4.

Your first example outputs 3. Your second outputs 4.

一般的经验法则是查找从最局部"到最不局部"变量.因此，这里的优先级是块 -> 本地 -> 类 -> 全局.

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.

您还可以显式访问每个阴影变量的大多数版本:

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();
}

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