我想实现 is_pointer。我想要这样的东西:
template <typename T >
bool is_pointer( T t )
{
// implementation
} // return true or false
int a;
char *c;
SomeClass sc;
someAnotherClass *sac;
is_pointer( a ); // return false
is_pointer( c ); // return true
is_pointer( sc ); // return false
is_pointer( sac ); // return true
我该如何实现? 谢谢
最佳答案
template <typename T>
struct is_pointer_type
{
enum { value = false };
};
template <typename T>
struct is_pointer_type<T*>
{
enum { value = true };
};
template <typename T>
bool is_pointer(const T&)
{
return is_pointer_type<T>::value;
}
约翰内斯指出:
This is actually missing specializations for T *const, T *volatile and T * const volatile i think.
解决方法:
template <typename T>
struct remove_const
{
typedef T type;
};
template <typename T>
struct remove_const<const T>
{
typedef T type;
};
template <typename T>
struct remove_volatile
{
typedef T type;
};
template <typename T>
struct remove_volatile<volatile T>
{
typedef T type;
};
template <typename T>
struct remove_cv : remove_const<typename remove_volatile<T>::type> {};
template <typename T>
struct is_unqualified_pointer
{
enum { value = false };
};
template <typename T>
struct is_unqualified_pointer<T*>
{
enum { value = true };
};
template <typename T>
struct is_pointer_type : is_unqualified_pointer<typename remove_cv<T>::type> {};
template <typename T>
bool is_pointer(const T&)
{
return is_pointer_type<T>::value;
}
...但当然,这或多或少只是重新发明了 std::type_traits
轮子:)
关于c++ - 如何实现is_pointer?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3177686/