[Bug c++/56693] New: Fail to ignore const qualification on top of a function type.

cassio.neri at gmail dot com gcc-bugzilla@gcc.gnu.org
Fri Mar 22 23:30:00 GMT 2013


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: unassigned@gcc.gnu.org
        ReportedBy: cassio.neri@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



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