c++ - 为什么这段代码不能用 VS2010 和 gcc 4.8.1 编译

Question

很简单，请看这段代码:

namespace B
{

    struct A{ int i; } ;

    A getA(int i);
}

//        ____ if I'll delete '::' then program successfull compiled.
//       /
::B::A  ::B::getA(int i){ ::B::A a = {i}; return a;}

#include <cstdio>
int main()
{

  ::B::A a = ::B::getA(2);

  printf("%d\n", a.i);

}

错误列表 VS2010:

1>main.cpp(94): error C3083: 'B': the symbol to the left of a '::' must be a type
1>main.cpp(94): error C2039: 'getA' : is not a member of 'B::A'
1>main.cpp(88) : see declaration of 'B::A'
1>error C4430: missing type specifier - int assumed. Note: C++ does not support default-int
1>main.cpp(94): error C2440: 'return' : cannot convert from 'B::A' to 'int'
1>          No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called
1>main.cpp(94): error C2617: 'getA' : inconsistent return statement
1>main.cpp(94) : see declaration of 'getA'

Gcc.4.8.1 错误列表(from ideone.com):

   prog.cpp:10:1: error: ‘B’ in ‘struct B::A’ does not name a type
 ::B::A  ::B::getA(int i){ ::B::A a = {i}; return a;}
 ^

问:这是一个错误还是我不明白什么？

Answer 1

一般而言， token 之间的空格没有任何意义，除非需要它来分隔 token 。所以这个:

::B::A  ::B::getA(...)

相当于

::B::A::B::getA(...)

要表明它们是两个独立的限定名，请在函数名两边使用括号:

::B::A (::B::getA)(...)

或者，如您所说，删除顶级限定符(尽管如果范围内还有其他名为 B 的东西，这可能会导致混淆)。