C++ - what type to supply for explicit instantiation of a template function, when std::less is one of the arguments?

Question

eg. Function header is:

template <class comparison_function, class iterator_type, class size_type>
bool F(comparison_function compare, iterator_type it)

Call takes iterator_type and comparison_function parameters, but not size_type. Hence the function call has to be explicit eg.:

F<?, std::vector<int>::iterator, long int>(std::less<int>(), it);

What would ? be?

Answer 1

What would ? be?

It should be the type of the first passed argument i.e., std::less<int>

//vvvvvvvvvvvvvv----------------------------------------vvvvvvvvvvvvvv
F<std::less<int>, std::vector<int>::iterator, long int>(std::less<int>(), it);

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Note that the parenthesis () in std::less<int>() means that you're passing a default constructed object of type std::less<int>.

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Sidenote

You can avoid passing two of the template arguments explicitly by making use of template argument deduction. This can be done by making size_type to be the first template parameter so that the remaining two template parameters can be deduced from the passed arguments as shown below:

//--------vvvvvvvvvvvvvvv--------------------------->now size_type is the first template parameter
template <class size_type, class comparison_function, class iterator_type>
void F(comparison_function compare, iterator_type it);

int main()
{
//----------vvvvvvvvvvvvvvvvvvvv---->this uses template argument deduction for the remaining two parameters 
    F<long>(std::less<int>(), it);
}