从函数类型中剥离所有限定符

Question

给定一个可能带有 cv-qualifier-seq 和 ref-qualifier 的可变参数函数类型，是否有可能写一个去除所有限定符的类型特征而不写 4 * 3 * 2 = 24 个偏特化？

template<class T>
struct strip_function_qualifiers;

template<class R, class... Args>
struct strip_function_qualifiers<R(Args...)> { using type = R(Args...); };

template<class R, class... Args>
struct strip_function_qualifiers<R(Args..., ...)> { using type = R(Args..., ...); };

template<class R, class... Args>
struct strip_function_qualifiers<R(Args...) const> { using type = R(Args...); };

template<class R, class... Args>
struct strip_function_qualifiers<R(Args..., ...) const > { using type = R(Args..., ...); };

template<class R, class... Args>
struct strip_function_qualifiers<R(Args...) const &> { using type = R(Args...); };

template<class R, class... Args>
struct strip_function_qualifiers<R(Args..., ...) const & > { using type = R(Args..., ...); };

// etc. etc. for each possible combination (24 in total)

并且 the new transactional memory TS 添加 transaction_safe ，这是否意味着我们需要为此编写 48 个部分专业化？

编辑：引用这些奇怪函数类型的描述（[dcl.fct]/p6，引用N4140）：

A function type with a cv-qualifier-seq or a ref-qualifier (including a type named by typedef-name (7.1.3, 14.1)) shall appear only as:

the function type for a non-static member function,

the function type to which a pointer to member refers,

the top-level function type of a function typedef declaration or alias-declaration,

the type-id in the default argument of a type-parameter (14.1), or

the type-id of a template-argument for a type-parameter (14.3.1).

[ Example:
typedef int FIC(int) const;
FIC f; // ill-formed: does not declare a member function
struct S {
   FIC f; // OK
};
FIC S::*pm = &S::f; // OK
— end example ]

The effect of a cv-qualifier-seq in a function declarator is not the same as adding cv-qualification on top of the function type. In the latter case, the cv-qualifiers are ignored. [Note: a function type that has a cv-qualifier-seq is not a cv-qualified type; there are no cv-qualified function types. — end note ] [ Example:
 typedef void F();
 struct S {
    const F f; // OK: equivalent to: void f();
 };
— end example ]

The return type, the parameter-type-list, the ref-qualifier, and the cv-qualifier-seq, but not the default arguments (8.3.6) or the exception specification (15.4), are part of the function type. [Note: Function types are checked during the assignments and initializations of pointers to functions, references to functions, and pointers to member functions. — end note ]

Answer 1

我看不出解决这个问题的方法 - 定义一次并尽可能重用它。
当限定符是 top-level 时，可以避免如此大量的特化 - 在这种情况下我们可以使用 std::remove_cv 或 std::remove_reference，删除每个中的所有正交限定符步。不幸的是，这不适用于您引用的段落中解释的功能：例如cv-qualifier 是函数类型的一部分，而不是顶级。 void() const 是与 void() 根本不同的类型，因此两者必须由两个不同的偏特化来匹配。

不过您可以使用宏来缩短所有专业：

#define REM_CTOR(...) __VA_ARGS__

#define SPEC(var, cv, ref) \
template <typename R, typename... Args> \
struct strip_function_qualifiers<R(Args... REM_CTOR var) cv ref > \
{using type = R(Args... REM_CTOR var);};

#define REF(var, cv) SPEC(var, cv,) SPEC(var, cv, &) SPEC(var, cv, &&)

#define CV(var) REF(var,) REF(var, const) \
                REF(var, volatile) REF(var, const volatile)

template <typename> struct strip_function_qualifiers;

CV(()) CV((,...))

Demo.
Boost.PP也是可以的：

#include <boost/preprocessor/tuple/enum.hpp>
#include <boost/preprocessor/seq/elem.hpp>
#include <boost/preprocessor/seq/for_each_product.hpp>

#define REF  (&&)(&)()
#define CV   (const volatile)(const)(volatile)()
#define VAR  (())((,...)) // Had to add a comma here and use rem_ctor below,
                          // otherwise Clang complains about ambiguous ellipses

#define SPEC(r, product) \
template <typename R, typename... Args> \
struct strip_function_qualifiers<R(Args... BOOST_PP_TUPLE_ENUM( \
    BOOST_PP_SEQ_ELEM(0, product))) \
    BOOST_PP_SEQ_ELEM(1, product)   \
    BOOST_PP_SEQ_ELEM(2, product)>  \
{using type = R(Args... BOOST_PP_TUPLE_ENUM(BOOST_PP_SEQ_ELEM(0, product)));};

template <typename> struct strip_function_qualifiers;

BOOST_PP_SEQ_FOR_EACH_PRODUCT(SPEC, (VAR)(CV)(REF))

Demo。添加新的限定符（例如 transaction_safe 或 transaction_safe_noinherit.

时，这两种方法都不会变得更长

这里是经过修改的 SPEC，它还定义了某些特征成员。

#include <type_traits>

#include <boost/preprocessor/tuple/size.hpp>

// […]

#define SPEC(r, product)                                         \
template <typename R, typename... Args>                          \
struct strip_function_qualifiers<R(Args... BOOST_PP_TUPLE_ENUM(  \
    BOOST_PP_SEQ_ELEM(0, product))) \
    BOOST_PP_SEQ_ELEM(1, product)   \
    BOOST_PP_SEQ_ELEM(2, product)>  \
{                                     \
    using type = R(Args... BOOST_PP_TUPLE_ENUM(BOOST_PP_SEQ_ELEM(0, product))); \
                                                            \
private:                                                    \
    using cv_type = int BOOST_PP_SEQ_ELEM(1, product);      \
    using ref_type = int BOOST_PP_SEQ_ELEM(2, product);     \
public:                                                     \
    using is_const    = std::is_const<cv_type>;             \
    using is_volatile = std::is_volatile<cv_type>;          \
    using is_ref_qualified = std::is_reference<ref_type>;               \
    using is_lvalue_ref_qualified = std::is_lvalue_reference<ref_type>; \
    using is_rvalue_ref_qualified = std::is_rvalue_reference<ref_type>; \
    using is_variadic = std::integral_constant<bool,                          \
                       !!BOOST_PP_TUPLE_SIZE(BOOST_PP_SEQ_ELEM(0, product))>; \
};

从函数类型中剥离所有限定符

Stripping all qualifiers from a function type

c++

templates

c++11

c++14

c++17