考虑以下类:

class MyClass
{
    int _id;
public:
    decltype(_id) getId();
};

decltype(MyClass::_id) MyClass::getId()
{
    return _id;
}

它编译得很好。

但是，当我用它制作模板类时:

template <class T>
class MyClass
{
    int _id;
public:
    decltype(_id) getId();
};

template <class T>
decltype(MyClass<T>::_id) MyClass<T>::getId()
{
    return _id;
}

我明白了:

test.cpp:10:27: error: prototype for 'decltype (MyClass<T>::_id) MyClass<T>::getId()' does not match any in class 'MyClass<T>'
 decltype(MyClass<T>::_id) MyClass<T>::getId()                                                                                
                           ^
test.cpp:6:19: error: candidate is: decltype (((MyClass<T>*)(void)0)->MyClass<T>::_id) MyClass<T>::getId()
     decltype(_id) getId();
                   ^

为什么会这样？
为什么会有不同的类型

• decltype (MyClass<T>::_id) MyClass<T>::getId()
• decltype (((MyClass<T>*)(void)0)->MyClass<T>::_id)

我可以通过在类中定义主体来修复它:

template <class T>
class MyClass
{
    int _id;
public:
    decltype(_id) getId() { return _id; }
};

尾随返回类型也有类似的问题:

template <class T>
class MyClass
{
    int _id;
public:
    auto getId() -> decltype(_id);
};

template <class T>
auto MyClass<T>::getId() -> decltype(MyClass<T>::_id)
{
    return _id;
}

错误:

test.cpp:10:6: error: prototype for 'decltype (MyClass<T>::_id) MyClass<T>::getId()' does not match any in class 'MyClass<T>'
 auto MyClass<T>::getId() -> decltype(MyClass<T>::_id)
      ^
test.cpp:6:10: error: candidate is: decltype (((MyClass<T>*)this)->MyClass<T>::_id) MyClass<T>::getId()
     auto getId() -> decltype(_id);
          ^

• decltype (MyClass<T>::_id) MyClass<T>::getId()
• decltype (((MyClass<T>*)this)->MyClass<T>::_id) MyClass<T>::getId()

g++ 5.3.0

最佳答案

根据标准草案N4582 §5.1.1/p13 一般 [expr.prim.general](强调我的):

An id-expression that denotes a non-static data member or non-static member function of a class can only be used:

(13.1) — as part of a class member access (5.2.5) in which the object expression refers to the member’s class⁶³ or a class derived from that class, or

(13.2) — to form a pointer to member (5.3.1), or

(13.3) — if that id-expression denotes a non-static data member and it appears in an unevaluated operand. [Example:

struct S {
int m;
};
int i = sizeof(S::m); // OK
int j = sizeof(S::m + 42); // OK

— end example ]

_{63) This also applies when the object expression is an implicit (*this) (9.3.1).}

同样来自 §7.1.6.2/p4 简单类型说明符 [dcl.type.simple](Emphasis Mine):

For an expression e, the type denoted by decltype(e) is defined as follows:

(4.1) — if e is an unparenthesized id-expression or an unparenthesized class member access (5.2.5), decltype(e) is the type of the entity named by e. If there is no such entity, or if e names a set of overloaded functions, the program is ill-formed;

(4.2) — otherwise, if e is an xvalue, decltype(e) is T&&, where T is the type of e;

(4.3) — otherwise, if e is an lvalue, decltype(e) is T&, where T is the type of e;

(4.4) — otherwise, decltype(e) is the type of e.

The operand of the decltype specifier is an unevaluated operand (Clause 5).

[Example:

const int&& foo();
int i;
struct A { double x; };
const A* a = new A();
decltype(foo()) x1 = 17; // type is const int&&
decltype(i) x2; // type is int
decltype(a->x) x3; // type is double
decltype((a->x)) x4 = x3; // type is const double&

— end example ] [ Note: The rules for determining types involving decltype(auto) are specified in 7.1.6.4. — end note ]

因此，由于 decltype 是未计算的操作数，因此代码是合法的，应该可以编译。

一个干净的解决方法是使用 decltype(auto):

template<typename T>
class MyClass {
  int _id;  
public:
  decltype(auto) getId();
};

template<typename T>
decltype(auto) MyClass<T>::getId() {
  return _id;
}

以上代码被 GCC/CLANG/VC++ 接受。

关于c++ - 原型(prototype)与 decltype 和 auto 不匹配，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/37185803/