function template has already been defined

[JUANDENT]

Hi,

I have

template<typename T> have_sort test_sort(decltype(&T::sort), decltype(&T::sort));
template<typename T> have_range test_sort(decltype(&T::begin), decltype(&T::end));

where:

struct have_sort
	{
		char c;
	};

struct have_range
{
	char d;
	have_sort  e;
};

with which i do:

template<typename T>
typename std::enable_if<sizeof(test_sort<T> 
                (NULL,NULL))==sizeof(have_sort)>::type fast_sort(T& x)
{
	x.sort();
}

template<typename T>
typename std::enable_if<sizeof(test_sort<T> 
               (NULL,NULL))==sizeof(have_range)>::type fast_sort(T& x)
{
	std::sort(x.begin(), x.end());
}

which should compile but I get the error:

std::enable_if<0,void>::type useEnableIf::fast_sort(T &)': function template has already been defined

why?

[Peter87]

If you pass NULL as argument to test_sort how is it supposed to know which overload to pick?

[JUANDENT]

enable_if should only permit instantiation of the correct function!!
should not clash!

[coder777]

Can you provide a compilable example?

What is the idea behind

std::enable_if<sizeof(test_sort<T> 
               (NULL,NULL))==sizeof(....)>::type

?

You may use std::is_same<..., ...>::value. See

http://www.cplusplus.com/reference/type_traits/is_same/

But is there any need for this? What is the type of T?

[Peter87]

JUANDENT, you might be right. My understanding of SFINAE is weak. But wouldn't it be a problem with std::list that has both begin and sort? Note that std::list::sort uses templates which seems to affect things, but I don't understand. Anyhow, my understanding is that you're not allowed to form pointers to standard library functions.

[JUANDENT]

yes Peter87 but that is another issue. For now lets not think about that...