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Subject: Re: [boost] [Fit] formal review ends 20th March.
From: paul Fultz (pfultz2_at_[hidden])
Date: 2016-03-22 20:33:35

On Tuesday, March 22, 2016 7:01 PM, Gavin Lambert <gavinl_at_[hidden]> wrote:

>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().

This doesn't really matter. `reference_wrapper` has a call operator that is
only const. Having a non-const overload for `reference_wrapper` would imply
that the function could change what `reference_wrapper` points to, but it
can't. So there is no reason to have a non-const overload. Same applies to
using `fit::indirect(&f)`.

> 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.)

With Fit, you can do `fit::indirect(std::make_shared<mutable_function>())` to
turn a shared_ptr that points to a Callable into a FunctionObject.
Additionally, if you absolutely know what you are doing, you can still use
`fit::mutable_` as well.
>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
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