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

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.