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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: Fri, 16 Aug 2024 11:00:54 +0100
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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++.  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?

As I said, I don't think it's wise to look at it this way, but I am just
borrowing your use of terms to try to tease out what you are getting at.

-- 
Ben.