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From: David Brown <david.brown@hesbynett.no>
Newsgroups: comp.lang.c
Subject: Re: "A diagram of C23 basic types"
Date: Wed, 9 Apr 2025 15:29:37 +0200
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On 08/04/2025 20:14, James Kuyper wrote:
> On 4/8/25 04:39, David Brown wrote:
>> On 07/04/2025 20:35, James Kuyper wrote:
>>> On 4/3/25 18:00, Waldek Hebisch wrote:
>>>> Tim Rentsch <tr.17687@z991.linuxsc.com> wrote:
>>> ...
>>>>> Not always practical.  A good example is the type size_t.  If a
>>>>> function takes an argument of type size_t, then the symbol size_t
>>>>> should be defined, no matter which header the function is being
>>>>> declared in.
>>>>
>>>> Why?  One can use a type without a name for such type.
>>>
>>> How would you declare a pointer to a function type such that it is
>>> compatible with such a function's type?
>>
>> The C23 "typeof" operator lets you work with the type of a value or
>> expression.  So you first have an object or value of type "size_t",
>> that's all you need.  Unfortunately, there are no convenient literal
>> suffixes that could be used here.
> 
> I can see how that would work with the return type of a function, but
> how would it apply to an argument of a function?
> 

Something like :

	memcpy(p, q, (typeof(sizeof(int))) 100);

I haven't tested that on a C23 compiler, because I really don't think it 
is the kind of thing I'd write in real code!