使用std :: remove_reference来获取STL容器的元素迭代器

Question

I'm experimenting with std::remove_reference. For example, I can extract an element type an array but how do I get remove_reference to work with STL containers? For example, I want to return an iterator to an element of a vector using remove_reference below:

#include <iostream>
#include <vector>
#include <type_traits>
using std::vector;
using std::cout;
using std::endl;
using std::begin;
using std::end;
using std::remove_reference;

template<typename T> 
auto my_end(T& c) -> typename remove_reference<decltype(&c[0])>::type
{
    return end(c)-1; //compile error when myend<vector> is instantiated
}

int main()
{
    int ia[] = {1,2,3,4,5,6,7,8,10};
    vector<int> v(begin(ia), end(ia));

    auto my_back1 = *my_end(ia);
    cout << my_back1 << endl; //prints 10

    auto my_back2 = *my_end(v);
    cout << my_back2 << endl; //should print 10
}

The compiler error when my_end<vector> is instantiated is:

cannot convert from 'std::_Vector_iterator<_Myvec>' to 'int *'  

Answer 1

What is the type returned by std::vector<T>::operator[] ? It is a T& . So, the result of decltype(&c[0]) is T* .

But what is the type of end(c)-1 ? It is an iterator.

If you want to return an iterator, use decltype(end(c)) or something similar.

---

Note that if you just want a reference to the last element, you can just use (or wrap):

ia.back();

and if you want an iterator (for some reason), but don't care about the direction :

ia.rbegin();

which will also de-reference to the last entry.

Your my_end is also unsafe if the container is empty ... but of course I'm not sure how you plan to use it.

Answer 2

What's the purpose of remove_reference here? std::vector<T>::end() yields an iterator, whereas &c[0] yields a pointer to an element.

Define what you are expecting my_end() to return, and then we can help you solve the problem.

If all you want is to return an iterator, then declare it like this:

auto my_end(T& c) -> decltype(begin(c)) { }

But then again, you would like to experiment with remove_reference , right...?

Answer 3

Thanks for your help folks. The code looks lame inline in my comment below so here it is again. This time remove_reference is useful to deduce the return type if for example I need to pass an iterator. (Okay so its not exactly that useful....)

#include <iostream>
#include <vector>
#include <type_traits>
using std::vector;
using std::cout;
using std::endl;
using std::begin;
using std::end;
using std::remove_reference;

template<typename T> 
auto my_end(T It) -> typename remove_reference<decltype(It)>::type
{
    return It-1; 
}

int main()
{
    int ia[] = {1,2,3,4,5,6,7,8,10};
    vector<int> v(begin(ia), end(ia));

    auto my_back1 = my_end(end(ia));
    cout << *my_back1 << endl; //prints 10

    auto my_back2 = my_end(end(v));
    cout << *my_back2 << endl; //should print 10
}