How to specialize Iterator by its value type, in C

Question

Is it possible to specialize an Iterator template parameter by its value_type?

I have a function with the following prototype.

template<typename InputIterator>
void f(InputIterator first, InputIterator last);

And I want to handle specially if InputIterator::value_type is SomeSpecificType.

Answer 1

Using SFINAE, assuming enable_if[_c] and is_same are either from Boost or <type_traits> (and are appropriately qualified with either boost:: or std:: respectively):

template<typename InputIterator>
typename enable_if<
    !is_same<
        typename std::iterator_traits<InputIterator>::value_type,
        SomeSpecificType
    >::value
>::type
f(InputIterator first, InputIterator last)
{
    // Default implementation.
}

template<typename InputIterator>
typename enable_if<
    is_same<
        typename std::iterator_traits<InputIterator>::value_type,
        SomeSpecificType
    >::value
>::type
f(InputIterator first, InputIterator last)
{
    // Special case
}

In the Boost case, use boost::enable_if_c for something similar to the above. You can use boost::enable_if and get rid of the ::value but then must also use e.g. boost::disable_if.

Answer 2

This will work. This type of specialization only works with structs, so I can't do this with a function.

template <typename InputIterator, typename ValueType = typename InputIterator::value_type>
struct foobar
{
    static void invoke(InputIterator first, InputIterator second)
    {
        // ...
    }
};

template <typename InputIterator>
struct foobar<InputIterator, SomeSpecificType>
{
    static void invoke(InputIterator first, InputIterator second)
    {
        // ...
    }
};

This shouldn't require you to set the type. It should be inferred automatically.

Answer 3

Call me naive, but why wouldn't the following suffice?

struct MyType; // the only type I want

#include <iterator>
typedef std::iterator<std::input_iterator_tag, MyType> MyIt;

void f(const MyIt & begin, const MyIt & end)
{
   /* ... */
}

---

OK, forget that above, that was nonsense. Here's a way to do it, which is just Luc's correct answer from above, for C++0x:

#include <vector>
#include <iterator>
#include <type_traits>

// "int" is our desired iterator value type, "void" is f's return type.
template <typename It>
typename std::enable_if<std::is_same<int, typename std::iterator_traits<It>::value_type>::value, void>::type
f(const It & begin, const It & end) { /* your function here */ }

int main()
{
  std::vector<double> x;
  std::vector<int> y;

  //f(x.cbegin(), x.cend()); // error
  f(y.cbegin(), y.cend());   // works
}

Answer 4

How about:

template<typename T>
typename std::enable_if<std::is_same<typename T::value_type, SomeType>::value, void>::type
f(T first, T second);

Answer 5

You can use some intermediate structs to get the partial template specialisation that you need. Something like this should do the trick

template<typename T, typename V>
struct f_impl
{
  static void f( T first, T last ) {...}; //Default version
};

template<typename T>
struct f_impl<T, SomeSpecificType>
{
   static void f(T first,T last) {...}; //Specialisation
};

template<typename InputIterator> void f(InputIterator first, InputIterator last)
{
  f_impl<
      InputIterator,
      typename std::iterator_traits<InputIterator>::value_type
  >::f(first,last);
};