Can't specialize formatter for templated ranged class

[drrlvn]

If I include fmt/ranges.h I cannot specialize fmt::formatter for my type that is ranged and templated.
Say I have this class:

template<typename T>
class Foo {
public:
    ...
    const_iterator begin() const { ... }
    const_iterator cbegin() const { ... }
    ...
};

When trying to provide a specialization of fmt::formatter for this class:

template<typename T>
struct formatter<Foo<T>> {
    ...
};

I get an ambiguous template instantiation error because of the specialization in fmt/ranges.h:

include/fmt/core.h:529:56: error: ambiguous template instantiation for ‘struct fmt::v5::formatter<Foo<int>, char, void>’
     typename Context::template formatter_type<T>::type f;
                                                        ^
In file included from ...:
include/fmt/ranges.h:199:8: note: candidates are: ‘template<class RangeT, class Char> struct fmt::v5::formatter<RangeT, Char, typename std::enable_if<fmt::v5::internal::is_range<T>::value>::type> [with RangeT = Foo<int>; Char = char]’
 struct formatter< RangeT, Char,
        ^~~~~~~~~~~~~~~~~~~~~~~~
     typename std::enable_if<fmt::internal::is_range<RangeT>::value>::type> {
     ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
In file included from ...:
foo.cpp: note:                 ‘template<class T> struct fmt::v5::formatter<Foo<T> > [with T = int]’
     struct formatter<Foo<T>> {
            ^~~~~~~~~~~~~~~~~

As a workaround I can force fmt::internal::is_range to be false:

namespace fmt::internal {

template<typename T>
struct is_range<Foo<T>> : std::false_type {};

} // namespace fmt::internal

But obviously I would want to avoid defining anything in an internal namespace.

[vitaut]

I think it's reasonable to make is_range part of the public API and move it to the fmt namespace. Could you submit a PR?

[drrlvn]

Sure, submitted #759.