如何获取延迟模板参数的类型

如果 TPtr 只能是指针 ,你在找 Boost's remove_pointer .

如果您想知道在这个工作的世界中是如何工作的,它会以这种方式使用部分专业化:

template<typename T>
struct RemovePointer;

template<typename T>
struct RemovePointer<T*>
{
    typedef T Type;
};

int main()
{
    RemovePointer<int*>::Type foobar; // foobar has the type int
    return 0;
}

如果 TPTR 可以是迭代器或指针 你需要 the iterator_traits class 是标准库的一部分。在您的情况下,它的用法如下:

template <class TPtr> 
struct foo { 
    typedef TPtr ptr_type;
    // The compiler won't know for sure if value_type is actually
    // a type until TPtr is known. The typename keyword is a hint
    // to the compiler so it doesn't cause an error.
    typedef typename iterator_traits<TPtr>::value_type element_type;
};

信不信由你,它也通过部分专业化运作。基本上是这样定义的:

// Primary template for iterators

template<class Iterator>
struct iterator_traits
{
    typedef typename Iterator::difference_type difference_type;
    typedef typename Iterator::value_type value_type;
    typedef typename Iterator::pointer pointer;
    typedef typename Iterator::reference reference;
    typedef typename Iterator::iterator_category iterator_category;
};

// Partial specializations for pointers

template<class T>
struct iterator_traits<T*>
{
    typedef ptrdiff_t difference_type;
    typedef T value_type;
    typedef T* pointer;
    typedef T& reference;
    typedef random_access_iterator_tag iterator_category;
};

template<class T>
struct iterator_traits<const T*>
{
    typedef ptrdiff_t difference_type;
    typedef T value_type;
    typedef const T* pointer;
    typedef const T& reference;
    typedef random_access_iterator_tag iterator_category;
};

这就是为什么 迭代器性状 类同时在迭代器和指针上工作。