How to pass const_iterator and non const iterator in a single template call

Question

I would like to know how to pass a const_iterator to a template. For example in the below call and function template.

The iterator for first1 and last 1 is

std::vector<std::pair<cv::Point_<float>, cv::Point_<float> > >::iterator

and the iterator for first2 and last2 is const_iterator

std::vector<std::pair<cv::Point_<float>, cv::Point_<float> > >::const_iterator

template < typename T>
T my_function(T __first1, T __last1, T __first2, T __last2, T __result)
{

}

Call is

my_function(co1.begin(), co1.end(), co2.begin(), co2,end(), result.begin());

Is there a way to pass both a normal iterator and a const_iterator in a single template?

The error is

error: no matching function for call to ‘my_function(std::vector<std::pair<cv::Point_<float>, cv::Point_<float> > >::iterator, std::vector<std::pair<cv::Point_<float>, cv::Point_<float> > >::iterator, std::vector<std::pair<cv::Point_<float>, cv::Point_<float> > >::const_iterator, std::vector<std::pair<cv::Point_<float>, cv::Point_<float> > >::const_iterator, std::vector<std::pair<cv::Point_<float>, cv::Point_<float> > >::iterator)’’
             my_function(co1.begin(), co1.end(), co2.begin(), co2.end(), result.begin());

I got it.. I just need to divide the template in T1 and T2..and it works ! template < typename T1, typename T2> T1 my_function(T1 __first1, T1 __last1, T2 __first2, T2 __last2, T1 __result) — infoclogged, Aug 02 '18 at 14:13
@user463035818 . found the answer myself.. should I delete this post? — infoclogged, Aug 02 '18 at 14:14
dont delete it when there are already answers given. Consider to accept one of them or if you like you should even post your own answer (more answers is always good) — 463035818_is_not_an_ai, Aug 02 '18 at 14:15
Just a unrelated comment for your interest: [Identifiers starting with two underscores are reserved for compiler vendors](https://stackoverflow.com/a/228797/1332041). — acraig5075, Aug 02 '18 at 14:20

LogicStuff · Accepted Answer · 2018-08-03T13:22:22.733

5

Obviously, you need different types so that deduction succeeds, not just T:

template <typename T1, typename T2>
T1 my_function(T1 __first1, T1 __last1, T2 __first2, T2 __last2, T1 __result) { ... }

You can add a SFINAE check (advanced!) (whichever way you want):

std::enable_if_t<std::is_convertible_v<T2, T1>> // C++17 way

to preserve the restriction that T2 is only a const version of T1. You might want to add a separate type for _result and the return type as well.

edited Aug 03 '18 at 13:22

answered Aug 02 '18 at 14:14

LogicStuff

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Most probably `result` also should have own type (`T3`). – Marek R Aug 02 '18 at 14:45

score 4 · Answer 2 · answered Aug 02 '18 at 14:14

Instead of trying to convert one iterator type to another just add another template type that will allow the two ranges to have two different types of iterators.

template < typename T, typename U, typename O>
T my_function(T __first1, T __last1, U __first2, U __last2, O __result)
{

}

^{added a third type for the output type as well}

This makes the function much more generic and flexible.

How to pass const_iterator and non const iterator in a single template call

2 Answers2