https://gcc.gnu.org/bugzilla/show_bug.cgi?id=90546
Bug ID: 90546
Summary: [9.1 regression] Incorrect template argument deduction
for conversion functions
Product: gcc
Version: 9.1.1
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c++
Assignee: unassigned at gcc dot gnu.org
Reporter: ndkrempel at gmail dot com
Target Milestone: ---
The following code deduces T = Foo using gcc 9.1, whereas gcc 8.3 (and clang 7
up) deduced T = const Foo. My reading of the standard (see below) is that T =
const Foo is correct.
struct Foo {};
void test(const Foo&);
struct Bar {
template <typename T> operator T&&();
};
int main() {
test(Bar{});
}
(compiled with -std=c++2a, but same behaviour with all language versions)
Relevant standards text: http://eel.is/c++draft/temp.deduct.conv
My reading of this applied to the above program is:
P is T&&, A is const Foo&
P is a reference type -> just treat as T from here on
A is not cv-qualified
A is a reference type -> use const Foo for type deduction
Now attempting to match T with const Foo succeeds with T = const Foo, so we
don't even need to consider the four bulleted exceptional cases ("These
alternatives are considered only if type deduction would otherwise fail.")
