命名空间内的友元函数声明/定义 [英] Friend function declaration/definition inside a namespace

问题描述

考虑一个命名空间内的类.类的定义声明了一个友元函数.

Consider a class inside a namespace. The definition of the class declares a friend function.

namespace Foo
{
    class Bar
    {
        friend void baz();
    };
}

据我所知，这应该将 baz() 声明为最内层封闭命名空间的成员，即 Foo.

This should, based on what I know, declare baz() as a member of the innermost enclosing namespace, i.e. Foo.

因此，我希望 baz() 的以下定义是正确的:

Therefore, I expected the following definition for baz() to be correct:

void Foo::baz() { }

但是，GCC (4.7) 给了我一个错误.

However, GCC (4.7) gives me an error.

error: ‘void Foo::baz()’ should have been declared inside ‘Foo’

几种解决方案似乎都有效:

Several solutions seem to work:

• 在类外声明 baz().

namespace Foo
{
    void baz();

    class Bar
    {
        friend void baz();
    };
}

在命名空间内定义baz().

namespace Foo
{
    class Bar
    {
        friend void baz();
    };
}
...
namespace Foo
{
    void baz() { }
}

使用 -ffriend-injection 标志编译，消除错误.

Compile with the -ffriend-injection flag, which eliminates the error.

这些解决方案似乎与我所知道的 C++ 中声明/定义的一般规则不一致.

These solutions seem to be inconsistent with the general rules of declaration/definition in C++ I know.

为什么我必须声明 baz() 两次?
为什么该定义仅在命名空间内是合法的，而在范围解析运算符中是非法的?
为什么flag会消除错误?

Why do I have to declare baz() twice?
Why is the definition otherwise only legal inside a namespace, and illegal with the scope resolution operator?
Why does the flag eliminate the error?

推荐答案

为什么我必须声明 baz() 两次?

Why do I have to declare baz() twice?

因为友元声明没有提供命名空间中函数的可用声明.它声明，如果在该命名空间中声明了该函数，它将是友元；如果您要在一个类中定义一个友元函数，那么它可以通过依赖于参数的查找(但不是其他方式)可用，就好像它是在命名空间中声明的一样.

Because the friend declaration doesn't provide a usable declaration of the function in the namespace. It declares that, if that function is declared in that namespace, it will be a friend; and if you were to define a friend function inside a class, then it would be available via argument-dependent lookup (but not otherwise) as if it were declared in the namespace.

为什么定义只在命名空间内是合法的，而在范围解析运算符中是非法的?

Why is the definition otherwise only legal inside a namespace, and illegal with the scope resolution operator?

因为它没有(正确地)在命名空间中声明，并且如果函数已经声明，则只能在其命名空间之外定义(使用范围解析).

Because it hasn't been (properly) declared in the namespace, and a function can only be defined outside its namespace (with scope resolution) if it has been declared.

为什么flag会消除错误?

Why does the flag eliminate the error?

因为标志导致友元声明充当命名空间中的声明.这是为了与 C++ 的古代方言(显然还有一些现代编译器)兼容，其中这是标准行为.

Because the flag causes the friend declaration to act as a declaration in the namespace. This is for compatibility with ancient dialects of C++ (and, apparently, some modern compilers) in which this was the standard behaviour.

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