c++ - 函数定义为限定 id

Question

为什么下面的代码编译成功:

#include <stdio.h>

namespace B
{
    void foo();
    int i=3;
}

void B::foo()
{
    i=65;
}
int main(){ }

以下未编译成功:

#include <stdio.h>

namespace B
{
    void foo()
    {
        i=65;//error: ‘i’ was not declared in this scope
    }
    int i=3;
}

int main(){ }

我想从目前澄清的标准中找到引用。

Answer 1

因为你想要引用 3.4.1 [basic.unqual.lookup]:

/4 A name used in global scope, outside of any function, class or user-declared namespace, shall be declared before its use in global scope.

/6 A name used in the definition of a function following the function’s declarator-id28 that is a member of namespace N (where, only for the purpose of exposition, N could represent the global scope) shall be declared before its use in the block in which it is used or in one of its enclosing blocks (6.3) or, shall be declared before its use in namespace N or, if N is a nested namespace, shall be declared before its use in one of N’s enclosing namespaces.