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Home/ Questions/Q 8745171
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Editorial Team
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Editorial Team
Asked: June 13, 20262026-06-13T11:57:46+00:00 2026-06-13T11:57:46+00:00

I’m experimenting with std::remove_reference. For example, I can extract an element type an array

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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 *'  
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  1. Editorial Team
    Editorial Team
    2026-06-13T11:57:47+00:00Added an answer on June 13, 2026 at 11:57 am

    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.

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