是否可以使用SFINAE /模板来检查运营商是否存在？

Question

我想检查运算符是否存在于编译时，如果它不是我只是想让它忽略，有没有办法做到这一点？

例如运营商：

template <typename T> 
QDataStream& operator<<(QDataStream& s, const QList<T>& l); 

Answer 1

我结束了使用备用命名空间：

namespace operators_fallback { 
template <typename T> 
inline QDataStream& operator<<(QDataStream& s, const T &) { return s; } 

template <typename T> 
inline QDataStream& operator>>(QDataStream& s, T &) { return s; } 

template <typename T> 
inline QDebug operator<<(QDebug d, const T &) { return d; } 
}; 

... 
inline void load(QDataStream & s) { 
    using namespace operator_fallback; 
    s >> item; 
} 

也发现了正确的方法来检查在编译时运营商（尽管我与回退会命名空间）。

或多或少基于此：

namespace private_impl { 
    typedef char yes; 
typedef char (&no)[2]; 

struct anyx { template <class T> anyx(const T &); }; 

no operator << (const anyx &, const anyx &); 
no operator >> (const anyx &, const anyx &); 


template <class T> yes check(T const&); 
no check(no); 

template <typename StreamType, typename T> 
struct has_loading_support { 
    static StreamType & stream; 
    static T & x; 
    static const bool value = sizeof(check(stream >> x)) == sizeof(yes); 
}; 

template <typename StreamType, typename T> 
struct has_saving_support { 
    static StreamType & stream; 
    static T & x; 
    static const bool value = sizeof(check(stream << x)) == sizeof(yes); 
}; 

template <typename StreamType, typename T> 
struct has_stream_operators { 
    static const bool can_load = has_loading_support<StreamType, T>::value; 
    static const bool can_save = has_saving_support<StreamType, T>::value; 
    static const bool value = can_load && can_save; 
}; 
} 
template<typename T> 
struct supports_qdatastream : private_impl::has_stream_operators<QDataStream, T> {}; 

template<typename T> 
struct can_load : private_impl::has_loading_support<QDataStream, T> {}; 

template<typename T> 
struct can_save : private_impl::has_saving_support<QDataStream, T> {}; 

template<typename T> 
struct can_debug : private_impl::has_saving_support<QDebug, T> {}; 

//编辑改变has_stream_operators了一下。

//编辑删除了链接，显然该网站有一些攻击的JavaScript。

Answer 2

这不是太容易，而且在C++ 03中它不是一般可能的。例如，如果您使用int*和int*作为op<<，则编译时您将遇到硬错误。因此对于非类别类型，您需要过滤掉标准禁止的类型。

对于op+我曾经为踢腿写过这样的东西。请注意，我用C头，因为我需要测试与clang编译的代码也一样，这在当时不支持我的C++头：

#include <stddef.h> 
#include <stdio.h> 

namespace detail { 
struct any { 
    template<typename T> any(T const&); 
}; 
struct tag { char c[2]; }; 

int operator,(detail::tag, detail::tag); 
template<typename T> void operator,(detail::tag, T const&); 
char operator,(int, detail::tag); 
} 

namespace fallback { 
    detail::tag operator+(detail::any const&, detail::any const&); 
} 

namespace detail { 
template<typename T> 
struct is_class { 
    typedef char yes[1]; 
    typedef char no[2]; 

    template<typename U> 
    static yes &check(int U::*); 
    template<typename U> 
    static no &check(...); 

    static bool const value = sizeof check<T>(0) == 1; 
}; 

template<typename T> 
struct is_pointer { typedef T pointee; static bool const value = false; }; 
template<typename T> 
struct is_pointer<T*> { typedef T pointee; static bool const value = true; }; 

template<typename T, typename U> 
struct is_same { 
    static bool const value = false; 
}; 

template<typename T> 
struct is_same<T, T> { 
    static bool const value = true; 
}; 

template<typename T> 
struct is_incomplete_array { 
    static bool const value = false; 
}; 

template<typename T> 
struct is_incomplete_array<T[]> { 
    static bool const value = true; 
}; 

template<typename T> 
struct is_reference { 
    typedef T referee; 
    static bool const value = false; 
}; 

template<typename T> 
struct is_reference<T&> { 
    typedef T referee; 
    static bool const value = true; 
}; 

// is_fn checks whether T is a function type 
template<typename T> 
struct is_fn { 
    typedef char yes[1]; 
    typedef char no[2]; 

    template<typename U> 
    static no &check(U(*)[1]); 

    template<typename U> 
    static yes &check(...); 

    // T not void, not class-type, not U[], U& and T[] invalid 
    // => T is function type 
    static bool const value = 
    !is_same<T const volatile, void>::value && 
    !is_class<T>::value && 
    !is_incomplete_array<T>::value && 
    !is_reference<T>::value && 
    (sizeof check<T>(0) == 1); 
}; 

template<typename T, bool = is_fn<T>::value> 
struct mod_ty { 
    typedef T type; 
}; 

template<typename T> 
struct mod_ty<T, true> { 
    typedef T *type; 
}; 

template<typename T> 
struct mod_ty<T[], false> { 
    typedef T *type; 
}; 

template<typename T, size_t N> 
struct mod_ty<T[N], false> { 
    typedef T *type; 
}; 

// Standard says about built-in +: 
// 
// For addition, either both operands shall have arithmetic or enumeration type, 
// or one operand shall be a pointer to a completely defined object type and 
// the other shall have integral or enumeration type. 

template<typename T> struct Ty; // one particular type 
struct P; // pointer 
struct Nc; // anything nonclass 
struct A; // anything 
struct Fn; // function pointer 

// matches category to type 
template<typename C, typename T, 
     bool = is_pointer<T>::value, 
     bool = !is_class<T>::value, 
     bool = is_fn<typename is_pointer<T>::pointee>::value> 
struct match { 
    static bool const value = false; 
}; 

// one particular type 
template<typename T, bool P, bool Nc, bool Fn> 
struct match<Ty<T const volatile>, T, P, Nc, Fn> { 
    static bool const value = false; 
}; 

// pointer 
template<typename T, bool F> 
struct match<P, T, true, true, F> { 
    static bool const value = true; 
}; 

// anything nonclass 
template<typename T, bool P, bool Fn> 
struct match<Nc, T, P, true, Fn> { 
    static bool const value = true; 
}; 

// anything 
template<typename T, bool P, bool Nc, bool Fn> 
struct match<A, T, P, Nc, Fn> { 
    static bool const value = true; 
}; 

// function pointer 
template<typename T> 
struct match<Fn, T, true, true, true> { 
    static bool const value = true; 
}; 

// one invalid combination 
template<typename A, typename B> 
struct inv; 

// a list of invalid combinations, terminated by B = void 
template<typename A, typename B> 
struct invs; 

// T[] <=> T[N] => T* 
// void() => void(*)() 
// T& => T 
// trying to find all invalid combinations 
// for built-in op+ 
typedef 
invs< 
    inv<Ty<float const volatile>, P>, 
invs< 
    inv<Ty<double const volatile>, P>, 
invs< 
    inv<Ty<long double const volatile>, P>, 
invs< 
    inv<Ty<void * const volatile>, Nc>, 
invs< 
    inv<Ty<void const* const volatile>, Nc>, 
invs< 
    inv<Ty<void volatile* const volatile>, Nc>, 
invs< 
    inv<Ty<void const volatile* const volatile>, Nc>, 
invs< 
    inv<Fn, Nc>, 
invs< 
    inv<Ty<void const volatile>, A>, 
invs< 
    inv<P, P>, 
void 
> > > > > > > > > > invalid_list; 

// match condition: only when ECond<true> is passed by specialization, 
// then it will be selected. 
template<bool> struct ECond; 

template<typename L, typename T, typename U, typename = ECond<true> > 
struct found_impl; 

// this one will first modify the input types to be plain pointers 
// instead of array or function types. 
template<typename L, typename T, typename U> 
struct found : found_impl<L, 
          typename mod_ty< 
          typename is_reference<T>::referee>::type, 
          typename mod_ty< 
          typename is_reference<U>::referee>::type> 
{ }; 

// match was found. 
template<typename F, typename B, typename R, typename T, typename U> 
struct found_impl<invs<inv<F, B>, R>, T, U, 
        ECond<(match<F, T>::value && match<B, U>::value) || 
         (match<B, T>::value && match<F, U>::value)> > { 
    static bool const value = true; 
}; 

// recurse (notice this is less specialized than the previous specialization) 
template<typename H, typename R, typename T, typename U, typename Ec> 
struct found_impl< invs<H, R>, T, U, Ec > : found_impl<R, T, U> { 
}; 

// we hit the end and found nothing 
template<typename T, typename U, typename Ec> 
struct found_impl< void, T, U, Ec > { 
    static bool const value = false; 
}; 

using namespace fallback; 

template<typename T, typename U, 
     bool found_invalid = found<invalid_list, T, U>::value> 
struct is_addable { 
    static T t; 
    static U u; 
    static bool const value = sizeof (detail::tag(), (t+u), detail::tag()) != 1; 
}; 

template<typename T, typename U> 
struct is_addable<T, U, true> { 
    static bool const value = false; 
}; 

} 

template<typename T, typename U> struct is_addable { 
    static bool const value = detail::is_addable<T, U>::value; 
}; 

当然，这样做的测试是非常重要的事后

// this one can be added 
struct test { 
    test operator+(test) { return(*this); } 
}; 

// this one cannot be added 
struct nono { }; 

// this fails because of an ambiguity, because there is a comma 
// operator taking a variable parameter on its left hand side.  
struct fails { fails operator+(fails); }; 

template<typename T> 
void operator,(T const&, fails); 


int main() { 
    printf("%d\n", is_addable<test, test>::value); 
    printf("%d\n", is_addable<int, float>::value); 
    printf("%d\n", is_addable<nono, nono>::value); 
    printf("%d\n", is_addable<int*, int>::value); 
    printf("%d\n", is_addable<int[1], int>::value); 
    printf("%d\n", is_addable<int[1], float[2]>::value); 
    printf("%d\n", is_addable<int*, float*>::value); 
    printf("%d\n", is_addable<void*, float>::value); 
    printf("%d\n", is_addable<void, int>::value); 
    printf("%d\n", is_addable<void(), int>::value); 
    printf("%d\n", is_addable<int, void(**)()>::value); 
    printf("%d\n", is_addable<float*&, int*&>::value); 
} 

Answer 3

这是一个老问题，但值得注意的只是增加这种能力对于几乎所有的运营商是加强与他们的最新Operator Type Traits。 OP询问的具体运营商是使用boost:has_left_shift进行测试。