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From: Ben Bacarisse <ben@bsb.me.uk>
Newsgroups: comp.lang.c
Subject: Re: technology discussion =?utf-8?Q?=E2=86=92?= does the world need
 a "new" C ?
Date: Mon, 19 Aug 2024 00:54:03 +0100
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David Brown <david.brown@hesbynett.no> writes:

> On 16/08/2024 12:00, Ben Bacarisse wrote:
>> David Brown <david.brown@hesbynett.no> writes:
>> 
>>> On 16/08/2024 02:08, Ben Bacarisse wrote:
>>>> Bart <bc@freeuk.com> writes:
>> 
>>>>> In general there is no reason, in a language with true call-by-reference,
>>>>> why any parameter type T (which has the form U*, a pointer to anything),
>>>>> cannot be passed by reference. It doesn't matter whether U is an array type
>>>>> or not.
>>>> I can't unravel this.  Take, as a concrete example, C++.  You can't pass
>>>> a pointer to function that takes an array passed by reference.  You can,
>>>> of course, pass a pointer by reference, but that is neither here nor
>>>> there.
>>>
>>> In C++, you can't pass arrays as parameters at all - the language inherited
>>> C's handling of arrays.  You can, of course, pass objects of std::array<>
>>> type by value or by reference, just like any other class types.
>> The best way to think about C++ (in my very non-expert opinion) is to
>> consider references as values that are passed by, err..., value.  But
>> you seem prepared to accept that some things can be "passed by reference"
>> in C++.  
>
> That seems a subtle distinction - I'll have to think about it a little. I
> like your description of arguments being like local variable initialisation
> - it makes sense equally well regardless of whether the parameter is "int",
> "int*", or "int&".  (It's probably best not to mention the other one in
> this group...)
>
>> So if this:
>> #include <iostream>
>> void g(int &i) { std::cout << i << "\n"; }
>> int main(void)
>> {
>>     int I{0};
>>     g(I);
>> }
>> shows an int object, I, being passed to g, why does this
>> #include <iostream>
>> void f(int (&ar)[10]) { std::cout << sizeof ar << "\n"; }
>> int main(void)
>> {
>>     int A[10];
>>     f(A);
>> }
>> not show an array, A, being passed to f?
>
> That's backwards compatibility with C array handling at play.

I'm not sure how this answers my question.  Maybe you weren't answering
it and were just making a remark...

-- 
Ben.