Can't specialize formatter for templated ranged class #751
Description
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::internalBut obviously I would want to avoid defining anything in an internal namespace.