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From: Bonita Montero <Bonita.Montero@gmail.com>
Newsgroups: comp.lang.c++,comp.lang.c
Subject: Re: Threads across programming languages
Date: Fri, 3 May 2024 17:18:55 +0200
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Am 03.05.2024 um 17:01 schrieb Michael S:
> On Fri, 3 May 2024 13:23:13 +0200
> Bonita Montero <Bonita.Montero@gmail.com> wrote:
> 
>> Am 03.05.2024 um 11:18 schrieb David Brown:
>>> On 03/05/2024 09:58, Bonita Montero wrote:
>>>> Am 03.05.2024 um 09:38 schrieb David Brown:
>>>>   
>>>>> No it is not.  C-style functions (or C++ functions for that
>>>>> matter) are not objects, and do not have calling operators.
>>>>> Built-in operators do not belong to a type, in the way that class
>>>>> operators do.
>>>>
>>>> You can assign a C-style function pointer to an auto
>>>> function-object.
>>>
>>> A C-style function /pointer/ is an object.  A C-style /function/ is
>>> not. Do you understand the difference?
>>
>> Practically there isn't a difference.
>>
> 
> For C, I agree, mostly because C has no nested functions.
> For C++ (after C++11) I am less sure, because of lambdas with
> non-empty captures.
> 

Lambdas without captures can be casted to C function-pointers and those
lambdas have all the same function-pointer type if the signature of the
calling operator is the same.
A nice trick to enforce function-pointer casting is to apply the +-ope-
rator to a non-capturing lambda since the plus-operator can be applied
to all pointers (I can really recommend the book "C++ Lambda Story" for
such details); this makes it possible to make the function-pointer defi-
nition type-inferenced. Or if you want to create a function<>-object
from the lambda which is guaranteed not to allocate if you pass a C
function-pointer you can enforce that if you attach the +-sign to the
assigned lambda.