using template parameters in function arguments does not work gcc4.8

Question

I have a function like this :

template<typename Iterator>
void sort2(Iterator it,
           std::function<bool(typename std::remove_pointer<
                              typename  Iterator::iterator_type>::type,
                                          int)> func)
{
}

int main()
{
    std::vector<int> a;
    sort2(a.begin(),[](int,int){return false;});
}
'main()::__lambda0' is not derived from 'std::function<bool(typename std::remove_pointer<typename Iterator::iterator_type>::type, int)>'
 sort2(a.begin(),[](int,int){return false;});
                                           ^

When I change it to :

template<typename Iterator>
void sort2(Iterator it,
           std::function<bool(typename std::remove_pointer<
                              typename  vector<int>::iterator/*I change this*/::iterator_type>::type,
                                          int)> func)
{
}

It compiles fine ...

What is wrong with the first function ?!

It seems correct ... why does it give compile errors ?

iterator::iterator_type will be a tag type (in this case, std::random_access). You probably just want Iterator::reference — IdeaHat
– IdeaHat, Commented May 27, 2014 at 13:47
Though I would point out that I would just template the second argument so that function pointers and operator structs could be used as well. — IdeaHat
– IdeaHat, Commented May 27, 2014 at 13:48
Note: passing a single iterator instead of a range will not work (although, it is an example) — user2249683
– user2249683, Commented May 27, 2014 at 13:53
@MadScienceDreams Iterator::iterator_type won't be anything - you're thinking of std::iterator_traits<Iterator>::iterator_category. — Casey
– Casey, Commented May 27, 2014 at 14:16

Casey · Accepted Answer · 2014-05-27 14:58:35Z

3

Iterators have no member named iterator_type. You want Iterator::value_type (Live at Coliru):

template<typename Iterator>
void sort2(Iterator it,
           std::function<bool(typename std::remove_pointer<
                              typename Iterator::value_type>::type,
                                          int)> func)
{
}

or better yet, use std::iterator_traits<Iterator>::value_type so your function will accept pointers as well as class type iterators (Coliru again):

template<typename Iterator>
void sort2(Iterator it,
           std::function<bool(typename std::remove_pointer<
                              typename std::iterator_traits<Iterator>::value_type>::type,
                                          int)> func)
{
}

A more generic version would accept the comparator type as a template parameter and use SFINAE to constrain that type to functions callable with two value_types having a return type convertible to bool instead of using convertibility to std::function<bool(value_type, value_type):

template <typename Iterator>
using ValueType = typename std::iterator_traits<Iterator>::value_type;

template<typename Iterator, typename Function>
auto sort2(Iterator first, Iterator last, Function func) ->
  typename std::enable_if<
    std::is_convertible<decltype(func(std::declval<ValueType<Iterator>>(),
                                      std::declval<ValueType<Iterator>>())),
                        bool>::value
  >::type
{
    std::sort(first, last, func);
}

It's hideously ugly - constraining with SFINAE always is - but at least gcc compiles it successfully.

edited May 27, 2014 at 14:58

answered May 27, 2014 at 14:11

Casey

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2 Comments

Praetorian Over a year ago

gcc still doesn't compile it. And iterator_type looked strange to me too, but I found some MSDN doc that talked about it.

Casey Over a year ago

@omid gcc compiles it if you force the second argument to be in a non-deduced context. So I suppose the question here is which compiler is wrong.

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