SFINAE - 尝试确定模板类型是否具有带有'variable'返回类型的成员函数 [英] SFINAE - Trying to determine if template type has member function with 'variable' return type

问题描述

使用SFINAE时遇到问题。我需要能够确定一个类型是否有一个成员函数operator-> defined，而不管它的返回类型。示例如下。

Having trouble with SFINAE. I need to be able to determine if a Type has a member function operator-> defined regardless of its return type. Example follows.

这个类在测试者中。它定义了operator - >（），返回类型为X *。因此，我不知道在任何地方都要用'X'来硬编码。

This class in the tester. It defines operator->() with a return type of X*. I thus will not know what 'X' is to hardcode it everywhere.

template <class X>
class PointerX
{
    ...

    X* operator->() const;
    ...
}

此类尝试确定传入的T有一个方法operator-> defined;而不管什么operator-> return类型。

This class tries to determines if the passed in T has a method operator-> defined; regardless of what operator-> return type is.

template<typename T>
struct HasOperatorMemberAccessor
{
    template <typename R, typename C> static R GetReturnType( R ( C::*)()const);

    template<typename U, typename R, R(U::*)()const> struct SFINAE{};
    template<typename U> static char Test(SFINAE<U,     decltype( GetReturnType(&U::operator->)),   &U::operator-> >*);
    template<typename U> static uint Test(...);
    static const bool value = sizeof(Test<T>(0)) == sizeof(char);
};

这个类与上面完全相同，除了operator-> return类型必须是Object 。

This class is the exact same as above except that operator-> return type has to be 'Object'.

template<typename T>
struct HasOperatorMemberAccessorOBJECT
{
    template <typename R, typename C> static R GetReturnType( R ( C::*)()const);

    template<typename U, typename R, R(U::*)()const> struct SFINAE{};
    template<typename U> static char Test(SFINAE<U,     Object*,                &U::operator-> >*); // only change is we hardcoded Object as return type.
    template<typename U> static uint Test(...);
    static const bool value = sizeof(Test<T>(0)) == sizeof(char);
};

结果：

void main()
{
    HasOperatorMemberAccessor<PointerX<Object>>::Test<PointerX<Object>>(0);         // fails  ::value is false;  Test => Test(...)

    HasOperatorMemberAccessorOBJECT<PointerX<Object>>::Test<PointerX<Object>>(0);       // works! ::value is true;   Test => Test(SFINAE<>*)  
}

HasOperatorMemberAccessor无法找到PointX的成员函数Object operator - >（）const。
所以它使用Test的通用版本Test（...）。

HasOperatorMemberAccessor was unable to find PointX's member function "Object operator->() const". So it uses Test's generic version Test(...).

但是，HasOperatorMemberAccessorOBJECT能够找到PointX的Object operator - >
因此它使用测试专用版本测试（SFINAE *）。

However, HasOperatorMemberAccessorOBJECT was able to find PointX's "Object operator->() const". Thus it uses Test specialized version Test(SFINAE*).

两个应该能够找到对象操作符 - >因此都应该使用Test的专业版Test（SFINAE *）;因此HasOperatorMemberAccessor> :: value应该为true。

Both should have been able to find the "Object operator->() const" method; and thus both should use Test's specialized version Test(SFINAE*); and thus HasOperatorMemberAccessor>::value should be true for both.

HasOperatorMemberAccessor和HasOperatorMemberAccessorOBJECT之间的唯一区别是HasOperatorMemberAccessorOBJECT具有对对象硬编码的类型名R，

The only difference between HasOperatorMemberAccessor and HasOperatorMemberAccessorOBJECT is that HasOperatorMemberAccessorOBJECT has the typename R hardcoded to object,

问题是decltype（GetReturnType（& U :: operator->））没有正确返回Object。我试过了一些不同的许可证发现的返回类型。它们如下：

So the issue is that "decltype( GetReturnType(&U::operator->))" is not returning Object correctly. I've tried a number of different permitations of discovering the return type. They go as follows:

    decltype( GetReturnType(&U::operator->) )
    typename decltype( GetReturnType(&U::operator->))
    decltype( ((U*)nullptr)->operator->() )
    typename decltype( ((U*)nullptr)->operator->() )

我正在使用MSVC ++ 10.0。

None work, why? I'm using MSVC++ 10.0.

推荐答案

你问的是如何实现这样的trait， decltype 不符合您的期望？如果是前者，这里有一种方法：

Are you asking how to implement such a trait, or why decltype isn't behaving as you expect? If the former, here's one approach:

#include <type_traits>

template<typename T, bool DisableB = std::is_fundamental<T>::value>
struct HasOperatorMemberAccessor
{ 
private:
    typedef char no;
    struct yes { no m[2]; };

    struct ambiguator { char* operator ->() { return nullptr; } };
    struct combined : T, ambiguator { };
    static combined* make();

    template<typename U, U> struct check_impl;
    template<typename U>
    static no check(
        U*,
        check_impl<char* (ambiguator::*)(), &U::operator ->>* = nullptr
    );
    static yes check(...);

public:
    static bool const value=std::is_same<decltype(check(make())), yes>::value;
};

// false for fundamental types, else the definition of combined will fail
template<typename T>
struct HasOperatorMemberAccessor<T, true> : std::false_type { };

// true for non-void pointers
template<typename T>
struct HasOperatorMemberAccessor<T*, false> :
    std::integral_constant<
        bool,
        !std::is_same<typename std::remove_cv<T>::type, void>::value
    >
{ };

template<typename X>
struct PointerX
{
    X* operator ->() const { return nullptr; }
};

struct X { };

int main()
{
    static_assert(
        HasOperatorMemberAccessor<PointerX<bool>>::value,
        "PointerX<> has operator->"
    );
    static_assert(
        !HasOperatorMemberAccessor<X>::value,
        "X has no operator->"
    );
    static_assert(
        HasOperatorMemberAccessor<int*>::value,
        "int* is dereferencable"
    );
    static_assert(
        !HasOperatorMemberAccessor<int>::value,
        "int is not dereferencable"
    );
    static_assert(
        !HasOperatorMemberAccessor<void*>::value,
        "void* is not dereferencable"
    );
}

VC ++ 2010缺少必要的C ++ 11设施使这更清洁。

VC++ 2010 lacks the necessary C++11 facilities (e.g. expression SFINAE) needed to make this much cleaner.

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