命名空间内的局部函数声明 [英] Local function declaration inside namespace

问题描述

在这种情况下

namespace n {
    void f() {
        void another_function();
    }
}

函数another_function应该在名称空间n内部还是外部定义? VS 2012(带有 11月CTP )说它应该在外面，而Mac上的GCC 4.7.2说它应该在里面.如果我做错了，我会从链接器中得到未定义的符号错误.

Should the function another_function be defined inside the namespace n or outside? VS 2012 (with the November CTP) says it should be outside, and GCC 4.7.2 on the Mac says it should be inside. If I do the wrong one, I get undefined symbol errors from the linkers.

我通常相信GCC更符合该标准，但这是C ++，您永远不能确定.

I generally trust GCC to be more compliant to the standard, but this is C++ and you can never be sure.

推荐答案

C ++ 11 3.5(以及C ++ 03)

C++11 3.5 (as well as C++03)

7 当未找到具有链接的实体的块范围声明时 引用其他声明，则该实体是的成员 最内层的封闭名称空间.但是，这样的声明并没有 在其名称空间范围内引入成员名称.

7 When a block scope declaration of an entity with linkage is not found to refer to some other declaration, then that entity is a member of the innermost enclosing namespace. However such a declaration does not introduce the member name in its namespace scope.

您的示例中的声明为n::another_function.

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