c++ - 我们是否需要在嵌套类型的成员参数前加上 "::"前缀？

Question

例如，考虑以下名称冲突的地方 nest1 :

template <typename U> class nest1 {};

class cls {
public:
    template <typename V> class nest1 {};

    template <typename W> class nest2 {
    public:
        void bar(nest1<W> x);
    };
};

template <typename W>
void cls::nest2<W>::bar(nest1<W> x) {}  // how does compiler know which nest<1>?

• 编译器如何知道bar是否存在？需要 nest1<W>或 cls::nest1<W>如果我们不使用 cls:: 作为前缀(例如 bar(cls::nest1<W> x))？
• 显式前缀 cls:: 是好习惯吗？无论如何？

注意:编译器实际上选择隐式声明 bar(cls::nest1<W> x) :

• 调用 cls::nest1<W> x; bar(x);作品:http://ideone.com/3ZuH2Z
• 路过nest1<W> x; bar(x);失败:http://ideone.com/6HmA3f

Answer 1

在成员函数名称之后使用的每个名称也在其类的词法范围内查找。这会导致以下(看似)不一致的行为，因为正常的返回类型不在类的词法范围内:

struct X{
  struct Y{};

  Y foo(Y);
  Y bar(Y);
};

// normal return type is before 'foo', needs explicit scope qualification
// parameter doesn't (after 'foo')
X::Y X::foo(Y y){ return y; }

// trailing-return-type also doesn't (after 'bar')
auto X::bar(Y y) -> Y{ return y; }

关于这方面的标准，我们看§9.3 [class.mfct] p5:

If the definition of a member function is lexically outside its class definition, the member function name shall be qualified by its class name using the :: operator. [ Note: A name used in a member function definition (that is, in the parameter-declaration-clause including the default arguments (8.3.6) or in the member function body) is looked up as described in 3.4. —end note ] [...]

然后在 §3.4.1 [basic.lookup.unqual] p8(非限定名称查找，例如没有 ::):

A name used in the definition of a member function (9.3) of class X following the function’s declarator-id³¹ [...] shall be declared in one of the following ways:

• [...]
• shall be a member of class X or be a member of a base class of X (10.2), or
• [...]

(在我的示例中，declarator-ids 是 foo 和 bar。)