Re: technology discussion → does the world need a "new" C ?

On 10/07/2024 02:18, Keith Thompson wrote:

bart <bc@freeuk.com> writes:

On 10/07/2024 00:50, Keith Thompson wrote:

bart <bc@freeuk.com> writes:
[...]

Arrays can be passed by explicit reference:
>
    void F(int(*A)[20]) {
        printf("%zu\n", sizeof(*A)/sizeof((*A)[0]));    // shows 20
    }
>
That can be called like this:
>
    int a[20];
    F(&a);

On the language level, that's passing a pointer to an array object.
The pointer itself is passed by value.  Passing a pointer to an array
is conceptually no different than passing a pointer to anything else.

>
I was replying to:
>
  "In C, arrays are not passed to functions, period."

 Which is a correct statement.
 [...]
 

But notice how C gives exactly the same result as my code that used
by-reference, even though:
>
  * C "doesn't pass arrays by reference"
  * C's F function uses the same parameter type (only & is missing; maybe
    by-ref is implicit...)
  * No explicit de-ref is needed inside F
  * No explicit address-of is needed when calling F

 Right.  The C rules that make all this possible have been explained
to you many times.  I won't waste my time explaining them to you
again.  If you were interested in learning, you would read section
6 of the comp.lang.c FAQ.
 Yes, some of C's rules make it *look like* arrays are passed by
reference.
 

So C behaves exactly as though it passes arrays by-reference, and yet
it doesn't have pass-by-reference. In fact, C does it without even
needing to be told!

 If you actually believed that C has pass-by-reference for arrays, it
would indicate that you don't understand C.  But you're only pretending
to believe it.
 If C had pass-by-reference for arrays, then presumably you could obtain
the size of an array parameter by applying sizeof to its name,

That's what my example did. But only if the array has a specific bound in the parameter type, not if it's unbounded, since (1) a function can be passed arrays of different sizes; (2) C arrays don't normally contain their length.

and you
could get the address of an array parameter by applying unary "&" to its
name.  I know why that doesn't work.  And so do you.

The by-ref in my language has a "&" has part of the parameter type; which would be cancelled by the "&" in "&a", so it would print the value of the passed pointer - the address of the array in the caller.
The C version doesn't need that "&" in the parameter (it can't be written), so it doesn't needthe "&" in "&a". Here then you just write "a" , and that gives the address of the array in the caller.
What I'm getting at, is that there is no appreciable difference between arrays passed by-reference in my language, and arrays as they are idiomatically passed in C.
So if arrays aren't passed by value in C, and they aren't passed by reference, then how the hell ARE they passed?! Pretend you have to give a quick answer to a child; or an alien.