Re: [boost] [result_of] const-qualified function?

From: Tobias Schwinger (tschwinger_at_[hidden])
Date: 2007-01-27 12:27:18

shunsuke wrote:
> Peter Dimov wrote:
>> shunsuke wrote:
>>> Hi, gurus.
>>>
>>> I have one question about 'result_of':
>>>
>>>
>>> template<class F>
>>> struct forward
>>> {
>>>    template<class A>
>>>    // typename boost::result_of<F(A&)>::type // (1)
>>>    typename boost::result_of<F const(A&)>::type // (2)
>>>    operator()(A& a) const
>>>    {
>>>        return m_f(a);
>>>    }
>>>
>>>    explicit forward(F f) : m_f(f)
>>>    { }
>>>
>>> private:
>>>    F m_f;
>>> };
>>>
>>>
>>> (1) or (2).. which is right?
>> (2) is right because m_f is const in operator().

...unless F is a reference.

>>
>>> In fact, (2) fails to compile if 'F' is a function pointer type;
>>> because const function-pointer specialization is missing.
>> This is a bug in boost::result_of.
> 
> Hmm, the result_of revolution is painful...

FunctionTypes provides a reimplementation of result_of (as one of its
examples) that can handle top-level cv-qualification correctly.

      http://tinyurl.com/qaf5f - zip archive

I replaced my local copy with it as boost::result_of was completely
broken, recently (renaming "namespace example" to "namespace boost") and
didn't experience any problems.

We can eventually replace boost::result_of with it, once FT is in and 
testing is back for the HEAD revision...

(BTW: the GCC configuration of FT has been fixed, so the problems 
reported during the review should be gone.)

Regards,
Tobias