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Home/ Questions/Q 6741567
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Editorial Team
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Editorial Team
Asked: May 26, 20262026-05-26T11:44:24+00:00 2026-05-26T11:44:24+00:00

I’d like to specialize std::iterator_traits<> for iterators of a container class template that does

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I’d like to specialize std::iterator_traits<> for iterators of a container class template that does not have the usual nested typedefs (like value_type, difference_type, etc.) and whose source I shouldn’t modify. Basically I’d like to do something like this:

template <typename T> struct iterator_traits<typename Container<T>::iterator> 
{
    typedef T value_type; 
    //  etc.
}; 

except that this doesn’t work, as the compiler is unable to deduce T from Container<T>::iterator.

Is there any working way to achieve the same?


For example:

template <typename T>
class SomeContainerFromAThirdPartyLib
{
    typedef T ValueType;    //  not value_type! 
    //  no difference_type

    class iterator
    {
        typedef T ValueType;    //  not value_type! 
        //  no difference_type  
        ...
    }; 
    iterator begin() { ... }
    iterator end() { ... }
    ...
}; 

Now suppose I call std::count() using an instance of this class. As far as I know, in most STL implementations, count() returns iterator_traits<Iterator>::difference_type. The primary template of iterator_traits<I> simply does typedef typename I::difference_type difference_type. Same with the other nested types.

Now in our example this obviously won’t work, as there’s no Container::iterator::difference_type. I thought I could work around this without modifying the iterator class, by specializing iterator_traits for iterators of any Container<T>.

In the end, I just want to be able to use std algorithms like count, find, sort, etc., preferably without modifying any existing code. I thought that the whole point of iterator_traits is exactly that: being able to specify types (like value_type, diff_type etc.) for iterator types that do not support them built-in. Unfortunately I can’t figure out how to specialize the traits class for all instances of Container<T>.

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T11:44:24+00:00Added an answer on May 26, 2026 at 11:44 am

    Yes. The compiler cannot deduce T from Container<T>::iterator because it is non-deducible context, which in other words means, given Container<T>::iterator, the value of T cannot uniquely and reliably be deduced (see this for detail explanation).

    The only solution to this problem is that you’ve to fully specialize iterator_traits for each possible value of iterator which you intend to use in your program. There is no generic solution, as you’re not allowed to edit the Container<T> class template.

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