http://gcc.gnu.org/bugzilla/show_bug.cgi?id=56693
Bug #: 56693
Summary: Fail to ignore const qualification on top of a
function type.
Classification: Unclassified
Product: gcc
Version: 4.8.0
Status: UNCONFIRMED
Severity: normal
Priority: P3
Component: c++
AssignedTo: unassig...@gcc.gnu.org
ReportedBy: cassio.n...@gmail.com
The following code:
void f() {}
template <typename T> void g(const T*) { }
int main() {
g(f);
}
raises an error with this note:
types 'const T' and 'void()' have incompatible cv-qualifiers
Attempting to instantiate g creates a function that takes a pointer to a const
T where T = void(). Since there's no such thing as a "const function", this
explains the note. However, C++11 8.3.5/6 says
"The effect of a cv-qualifier-seq in a function declarator is not the same as
adding cv-qualification on top of the function type. In the latter case, the
cv-qualifiers are ignored."
Hence, the const qualifier should be ignored and the code should compile. (It
does compile with clang and visual studio.)
For more information see:
http://stackoverflow.com/questions/15578298/can-a-const-t-match-a-pointer-to-free-function
