Re: [boost] [Fit] formal review ends 20th March.

Subject: Re: [boost] [Fit] formal review ends 20th March.
From: Gavin Lambert (gavinl_at_[hidden])
Date: 2016-03-22 20:00:32

On 22/03/2016 06:55, Paul Fultz II wrote:
>> Function Objects
>>
>> You state that the function call operator is always declared const.  On
>> what basis do you make that claim?
>
> This I plan to discuss more. There are two main reason for this:
>
> 1) Mutable function objects are problematic with many surprises. Most of
> the time `std::ref` should be used instead.

These are not exclusive.  I've often created a "real" function object 
(custom struct/class) with non-const call operator, and then used 
bind(ref(func), params...) on it.  This works because bind can call both 
const and non-const functors and can auto-unwrap ref().  The output of 
bind() might be a const functor, but its input is not required to be.

(And in some cases using ref() is not possible, as the functor must be 
returned out of the original scope.  Admittedly this is harder to do 
correctly with non-const functors than with const ones, but shared_ptr 
and friends can come to the rescue here, albeit with some minor 
performance cost.)

As I mentioned in the other branch of this thread, my experience has 
been that any time I want a const functor I use bind() or a lambda, and 
any time I want a non-const functor I write a custom struct.  Thus, the 
only times I actually implement operator() myself, it is never const.  I 
don't know how wide-spread this practice is, but I don't believe it is 
unique.