在命名空间中的Friend函数声明/定义 [英] Friend function declaration/definition inside a namespace

问题描述

考虑一个命名空间中的类。类的定义声明一个朋友函数。

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

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

这应该基于我所知道的，声明 baz（）作为最内层命名空间的成员，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() { }

。

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

几种解决方案似乎有效：

Several solutions seem to work:

• 在类外部声明 baz（）。

namespace Foo
{
    void baz();

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

定义 baz c $ c>命名空间中。

Define baz() inside the namespace.

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

使用 -ffriend-injection 标志进行编译， >

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

这些解决方案似乎与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？

为什么定义在命名空间中是合法的，

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?

因为朋友声明不提供可用的声明的命名空间中的函数。它声明，如果该函数在该命名空间中声明，它将是一个朋友;如果您要在类中定义友元函数，那么它将通过参数相关查找（但不是其他方式），就像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.

为什么这个定义在命名空间中是合法的，并且是范围解析操作符的非法？

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

因为它没有（正确）在命名空间中声明，并且只有在已声明的情况下，才能在其命名空间之外定义一个函数（具有范围解析）。

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.

为什么标志消除了错误？

Why does the flag eliminate the error?

朋友声明作为命名空间中的声明。这是为了与古代方言的C ++（和显然，一些现代编译器）的兼容性，其中这是标准的行为。

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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