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

On 7/10/24 01:55, Lawrence D'Oliveiro wrote:

On Sat, 06 Jul 2024 15:23:47 -0700, Keith Thompson wrote:
 

Lawrence D'Oliveiro <ldo@nz.invalid> writes:
>

On Fri, 05 Jul 2024 11:46:38 -0700, Keith Thompson wrote:
>

No, arrays are not pointers.

>
Except array indexing is designed to be indistinguishable from pointer
arithmetic.

>
No, arrays are not pointers.

 
Can you point out any situation where this construct
 
    &a[b]
 
might be valid, but this
 
    (a + b)
 
(with the same declarations of “a” and “b”) might not?

&a[b] is defined as equivalent to &*(a+b), and which in turn is
equivalent to (a+b), so of course that's not a problem. The fact that an
array is not a pointer is not visible in that context, but is visible in
several others:

#include <stdio.h>

int main(void)
{
    char *pointer = "array 1";
    char array[] = "array 2";
    // "array 2" is an lvalue of array type, and in most contexts,
    // gets implicitly converted into a pointer to the first character
    // of that array, in this context it does not, and this:
    // char array[] = pointer;
    // would therefore be a constraint violation;

    //These aren't guaranteed to be unequal, but they are allowed to
    // be, and are unlikely to be equal on most implementations.
    printf("%zu != %zu", sizeof array, sizeof pointer);
    // Even if the types did have the same size, you can demonstrate
    // that they are different using _Generic()


    // The following declarations declare objects of different types.
    typeof(array) newarray;
    typeof(pointer) newpointer;

    printf("%zu != %zu", sizeof array, sizeof pointer);

    // The following two statements would be constraint violations
    // if you swapped "pointer" with "array":
    char **ppchar = &pointer;
    char (*parray)[8] = &array;
}