在C 03中,如何确定类型T是否可解除引用？
我的意思是,我如何静态地确定* t是否是T类型t的有效表达式？
我的意思是,我如何静态地确定* t是否是T类型t的有效表达式？
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)上崩溃和灼伤.
解决方法
#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 >
{ };
我在msvs 2008中测试过它
