如何确定C ++ 03中是否可取消引用类型？ [英] How to determine if a type is dereferenceable in C++03?

问题描述

在 C ++ 03 中，如何确定类型 T 是否可取消引用？

意思是，我如何静态确定 * t 是否是类型 t 的有效表达式> T ？

In C++03, how do I determine if a type T is dereferenceable?
By which I mean, how do I statically determine if *t would be a valid expression for t of type T?

我的尝试：

template<bool B, class T = void> struct enable_if { };
template<class T> struct enable_if<true, T> { typedef T type; };

unsigned char (&helper(void const *))[2];
template<class T>
typename enable_if<
    !!sizeof(**static_cast<T *>(NULL)),
    unsigned char
>::type helper(T *);

template<class T>
struct is_dereferenceable
{ static bool const value = sizeof(helper(static_cast<T *>(NULL))) == 1; };

struct Test
{
    int *operator *();
    void operator *() const;
private:
    Test(Test const &);
};

int main()
{
    std::cout << is_dereferenceable<int *>::value;       // should be true
    std::cout << is_dereferenceable<void *>::value;      // should be false
    std::cout << is_dereferenceable<Test>::value;        // should be true
    std::cout << is_dereferenceable<Test const>::value;  // should be false
}

它适用于GCC（打印 1010 ），但在VC ++（ 1110 ）和Clang（ 1111 ）上崩溃并烧毁。

It works on GCC (prints 1010) but crashes and burns on VC++ (1110) and Clang (1111).

推荐答案

#include <boost\type_traits\remove_cv.hpp>
#include <boost\type_traits\is_same.hpp>
#include <boost\type_traits\remove_pointer.hpp>
#include <boost\type_traits\is_arithmetic.hpp>
#include <boost\utility\enable_if.hpp>

namespace detail
{
    struct tag 
    { 
        template < typename _T > 
        tag(const _T &); 
    };

    // This operator will be used if there is no 'real' operator
    tag operator*(const tag &);

    // This is need in case of operator * return type is void
    tag operator,(tag, int);  

    unsigned char (&helper(tag))[2];

    template < typename _T >
    unsigned char helper(const _T &);

    template < typename _T, typename _Enable = void >
    struct traits
    {
        static const bool value = (sizeof(helper(((**static_cast <_T*>(NULL)), 0))) == 1);
    };

    // specialization for void pointers
    template < typename _T >
    struct traits < _T,
        typename boost::enable_if < typename boost::is_same < typename boost::remove_cv < typename boost::remove_pointer < _T > ::type > ::type, void > > ::type >
    {
        static const bool value = false;
    };

    // specialization for arithmetic types
    template < typename _T >
    struct traits < _T,
        typename boost::enable_if < typename boost::is_arithmetic < typename boost::remove_cv < _T > ::type > > ::type >
    {
        static const bool value = false;
    };
}

template < typename _T > 
struct is_dereferenceable :
    public detail::traits < _T >
{ };

我已经在msvs 2008中对其进行过测试

I have tested it in msvs 2008

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