为泛型函数定义自定义迭代器特征

Question

I'm trying to write a generic function which will derive a return type at compile time according to the iterator it is given. Usually this is done through std::iterator_traits, but I also wanted to define my own version of iterator_traits, called my_iterator traits. Here's my code:

#include <cassert>
#include <iterator>
#include <vector>

using namespace std;

template <typename I>
struct my_iterator_traits {
    typedef typename I::value_type                     value_type;
    typedef typename I::iterator_category       iterator_category;
};

template <typename T>
struct my_iterator_traits<T*> {
    typedef T                                         value_type;
    typedef std::random_access_iterator_tag    iterator_category;
};                                                               

template <typename II>
typename my_iterator_traits<II>::value_type f(II b, II e) {
    typename my_iterator_traits<II>::value_type val = 0;
    return val;
}

int main () {
    int a[] = {2, 3, 4};
    int i = f(a, a + 3);

    assert(i == 0);

    // vector<int> v = {2};
    // f(v.begin(), v.end());

    // assert(j == 0);
    return 0;
}

Everything up to and including the main function and the function f() makes perfect sense. In main, I make an int array, call f() on it, and confirm that the output I get is an int whose value is zero.

What isn't clear is the following. I have two templates up top followed by the struct keyword. It would make perfect sense to use the second one (the one that takes in T* as the template argument). Specifically, why do we need the first struct template (the one without template parameters)? More importantly, what is the relationship between the two struct templates?

Answer 1

The first is the declaration of the template and the second is a partial specialization of the first. Partial specialization like explained in the link can have customization of the template parameter (ex: some fixed parameters, some restriction of what parameter your are referring (pointer, references, etc...), etc...).

Answer 2

The first template<typename>struct is a template , the second is a specialization of the first template .

template specialization is something similar to overriding (but not really) that is based off of pattern matching. If the arguments to the template can be matched by the pattern of a specialization, it is used instead.

This pattern matching does not do type conversion in the same way that function overrides do. SFINAE is in play however, for advanced work.