通过传递输出迭代器从函数填充std :: [container]

Question

I want to populate a container from inside a function by passing an output iterator as this is the most efficient way to do it as I understand. eg

template <typename OutputIterator>
void getInts(OutputIterator it)
{
   for (int i = 0; i < 5; ++i)
     *it++ = i;
}

( Is returning a std::list costly? )

But how can I enforce the type, the iterator should point to ? Basically I want to say "this function takes an output iterator of type boost::tuple" .

Answer 1

You can use boost::enable_if in conjunction with std:iterator_traits :

#include <boost/type_traits/is_same.hpp>
#include <boost/utility/enable_if.hpp>

template <typename OutputIterator>
typename boost::enable_if<
    boost::is_same<
        int, /* replace by your type here */
        typename std::iterator_traits<OutputIterator>::value_type
    >
>::type getInts(OutputIterator it)
{
   for (int i = 0; i < 5; ++i)
     *it++ = i;
}

Answer 2

You don't need to. The code will fail to compile anyway if the caller passes the wrong iterator type.

So it's enforced for you already.