Comments (6)
viboes
commented on June 2, 2024
There is a missing
using strong_typedef::strong_typedef;
inside X.
The same for Y.
from type_safe.
JohelEGP
commented on June 2, 2024
The compiler-generated default constructors default construct the bases, so that's not the problem.
Here's the actual error:
main.cpp:6:22: error: use of overloaded operator '==' is ambiguous (with operand types 'Y' and 'Y')
constexpr bool _{Y{} == Y{}}; // errors
~~~ ^ ~~~
type_safe/strong_typedef.hpp:310:9: note: candidate function
[with StrongTypedef = X, Other = Y, $2 = void]
TYPE_SAFE_DETAIL_MAKE_OP(==, equality_comparison, bool)
^
type_safe/strong_typedef.hpp:206:22: note: expanded from macro 'TYPE_SAFE_DETAIL_MAKE_OP'
constexpr Result operator Op(Name<StrongTypedef>&& lhs, Other&& rhs) \
^
type_safe/strong_typedef.hpp:310:9: note: candidate function
[with StrongTypedef = X, Other = Y, $2 = void]
type_safe/strong_typedef.hpp:198:22: note: expanded from macro 'TYPE_SAFE_DETAIL_MAKE_OP'
constexpr Result operator Op(const Name<StrongTypedef>& lhs, Other&& rhs) \
^
type_safe/strong_typedef.hpp:310:9: note: candidate function
[with StrongTypedef = X, Other = Y, $2 = void]
type_safe/strong_typedef.hpp:223:22: note: expanded from macro 'TYPE_SAFE_DETAIL_MAKE_OP'
constexpr Result operator Op(Other&& lhs, Name<StrongTypedef>&& rhs) \
^
type_safe/strong_typedef.hpp:310:9: note: candidate function
[with StrongTypedef = X, Other = Y, $2 = void]
type_safe/strong_typedef.hpp:214:22: note: expanded from macro 'TYPE_SAFE_DETAIL_MAKE_OP'
constexpr Result operator Op(Other&& lhs, const Name<StrongTypedef>& rhs) \
^
1 error generated.
from type_safe.
foonathan
commented on June 2, 2024
Thanks for reporting this. I've now added test cases to prevent further errors.
Note that you're example still doesn't compile under C++11 due to constexpr limitations I can't workaround.
from type_safe.
JohelEGP
commented on June 2, 2024
I'm still getting the same error (with the added overloads).
I think I found out why this is happening from looking at the following error from my non-contrived, still unreleased source code.
../include/rpg2d/eng/utility.hpp:60:16: error: use of overloaded operator '==' is ambiguous (with operand types 'rpg2d::eng::Abscissa'
(aka 'Unit<int, struct _abscissa>') and 'rpg2d::eng::Abscissa')
return l.x == r.x && l.y == r.y;
~~~ ^ ~~~
type_safe/strong_typedef.hpp:342:9: note: candidate function [with StrongTypedef =
jegp::Strong_typedef<int, rpg2d::eng::_abscissa>, Other = rpg2d::Unit<int, rpg2d::eng::_abscissa> &, $2 = void]
TYPE_SAFE_DETAIL_MAKE_OP(==, equality_comparison, bool)
^
type_safe/strong_typedef.hpp:230:22: note: expanded from macro 'TYPE_SAFE_DETAIL_MAKE_OP'
constexpr Result operator Op(const Name<StrongTypedef>& lhs, Other&& rhs) \
^
type_safe/strong_typedef.hpp:342:9: note: candidate function [with StrongTypedef =
jegp::Strong_typedef<int, rpg2d::eng::_abscissa>, Other = rpg2d::Unit<int, rpg2d::eng::_abscissa> &, $2 = void]
type_safe/strong_typedef.hpp:246:22: note: expanded from macro 'TYPE_SAFE_DETAIL_MAKE_OP'
constexpr Result operator Op(Other&& lhs, const Name<StrongTypedef>& rhs) \
^
type_safe/strong_typedef.hpp:342:9: note: candidate function [with StrongTypedef =
jegp::Strong_typedef<int, rpg2d::eng::_abscissa>]
type_safe/strong_typedef.hpp:198:22: note: expanded from macro 'TYPE_SAFE_DETAIL_MAKE_OP'
constexpr Result operator Op(const Name<StrongTypedef>& lhs, const Name<StrongTypedef>& rhs) \
^Unit is a Strong_typedef (which is a type_safe::strong_typedef) by public inheritance. The new overload (last candidate in the error) is correctly considered, but the ones with the Other parameter are too. This is the SFINAE condition of the Other overloads:
template <typename From, typename To>
using enable_if_convertible = typename std::enable_if<
!std::is_same<typename std::decay<From>::type, To>::value
&& std::is_convertible<typename std::decay<From>::type, To>::value>::type;From is Other&& and To the StrongTypedef. So in my error From is Unit and To is Strong_typedef. As Unit is convertible to Strong_typedef, the Other overloads are considered. Simply adding !std::is_base_of<To, typename std::decay<From>::type>::value to the SFINAE check considers this composition by inheritance and makes the code well-formed.
from type_safe.
foonathan
commented on June 2, 2024
I'm sorry, I modified your example and forgot to change it back later.
This was the only test case, I'm missing you're right and your fix works.
from type_safe.
JohelEGP
commented on June 2, 2024
Thanks a bunch! :)
from type_safe.
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from type_safe.