default value for template function, functional parameter

Question

template<typename Iterator, typename typename Comparator = std::less<typename std::iterator_traits<Iterator>::value_type>>
static void sort(Iterator begin, Iterator end, Comparator cmp = Comparator())
{
    ...
}

I have the following template function:

template<typename func>
static void sort_test(func sort)
{
    ...
    sort(somevector.begin(), somevector.end());
    ...
}

int main()
{
    sort_test(&sort<vector<int, allocator<int>>::iterator>);
    return 0;
}

error C2198: 'void (__cdecl *)(iterator,iterator,std::less)' : too few arguments for call

If I try to bypass the default argument by providing it:

template<typename func>
static void sort_test(func sort)
{
    ...
    sort(somevector.begin(), somevector.end(), std::less<int>);
    ...
}

int main()
{
    sort_test(&sort<vector<int, allocator<int>>::iterator>,
                    std::less<int>);
    return 0;
}

error C2275: 'std::less' : illegal use of this type as an expression

Praetorian · Accepted Answer · 2015-04-28 23:54:56Z

2

Default arguments are not part of a function's signature, so when you go through a pointer to function indirection, as you're doing in the first example, that information is lost. If you call your sort function directly within main with 2 iterator arguments, the code will compile.

In the second example you're getting a compilation error because you're trying to pass a type, instead of an instance, to sort

sort(somevector.begin(), somevector.end(), std::less<int>());

You also have an extra typename in the template parameter list for sort

template<typename Iterator, typename typename Comparator = std::less<typename std::iterator_traits<Iterator>::value_type>>
//                          ^^^^^^^^

answered Apr 28, 2015 at 23:54

Praetorian

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