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From: "potswa at mac dot com" <gcc-bugzilla@gcc.gnu.org>
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Subject: [Bug libstdc++/56283] New: std::result_of does not gracefully SFINAE (not define ::type)
Date: Mon, 11 Feb 2013 04:10:00 -0000
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http://gcc.gnu.org/bugzilla/show_bug.cgi?id=56283

             Bug #: 56283
           Summary: std::result_of does not gracefully SFINAE (not define
                    ::type)
    Classification: Unclassified
           Product: gcc
           Version: 4.7.2
            Status: UNCONFIRMED
          Severity: normal
          Priority: P3
         Component: libstdc++
        AssignedTo: unassigned@gcc.gnu.org
        ReportedBy: potswa@mac.com


>>From the spec of std::result_of,

If the expression INVOKE(declval<Fn>(), declval<ArgTypes>()...) is well formed
when treated as an unevaluated operand (Clause 5), the member typedef type
shall name the type decltype(INVOKE(declval<Fn>(), declval<ArgTypes>()...));
otherwise, there shall be no member type.

If the given INVOKE expression is invalid, ::type should not exist. But an
attempt to use it this way to detect a valid call fails.

Below is a short demo program. A non-overloaded function name and a functor
type with an overloaded call operator are passed to result_of through a type
traits style SFINAE wrapper. None of the cases work. In 2 of the 3 cases the
failure fires the static assertion, but not for the multiple overload issue.


#include <type_traits>

template< typename fn, typename = void >
struct is_invokable : std::false_type {};

template< typename fn >
struct is_invokable< fn, typename std::conditional< true, void, typename
std::result_of< fn >::type >::type >
        : std::true_type {};

void f( int );
static_assert( ! is_invokable< decltype( f ) &( char const * ) >(), "can
convert c-string to int" );

struct s {
    void operator() ( int ) {}
    void operator() ( float ) {}
};
static_assert( ! is_invokable< s( double ) >(), "resolved ambiguous conversion"
);
static_assert( ! is_invokable< s( char * ) >(), "converted pointer to number"
);