我正在尝试检查编译时是否存在运算符，如果不存在，我只想忽略它，有什么办法吗？
示例运算符：
template <typename T>
QDataStream& operator<<(QDataStream& s, const QList<T>& l);
问问题
4595 次
3 回答
12
我最终使用了一个后备命名空间：
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;
}
还找到了在编译时检查运算符的正确方法（尽管我将使用后备命名空间）。
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> {};
//edit 稍微改变了 has_stream_operators。
//edit删除了链接，显然该网站有一些攻击javascript。
于 2010-12-14T00:07:26.753 回答
6
这是一个老问题，但值得注意的是，Boost 刚刚为几乎所有具有最新Operator Type Traits的 Operator 添加了此功能。询问的特定运算符 OP 已使用boost:has_left_shift.
于 2012-01-12T03:03:28.583 回答
2
这不太容易，在 C++03 中通常是不可能的。例如，如果您使用int*and ，您将在编译时遇到硬错误。所以对于非类类型，你需要过滤掉标准禁止的类型。int*op<<
因为op+我曾经写过这样的东西。请注意，我使用的是C标头，因为我还需要使用clang编译器测试代码，当时它不支持我的 C++ 标头：
#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，事后做测试很重要
// 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);
}
于 2010-12-16T05:09:33.813 回答