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If the code and the comments disagree, then both are probably wrong. -- Norm Schryer

devel / comp.lang.c / Re: I think references should have been const by default

Re: I think references should have been const by default

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https://www.novabbs.com/devel/article-flat.php?id=18749&group=comp.lang.c#18749

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From: jameskuy...@alumni.caltech.edu (James Kuyper)
Newsgroups: comp.lang.c
Subject: Re: I think references should have been const by default
Date: Tue, 26 Oct 2021 14:32:06 -0400
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by: James Kuyper - Tue, 26 Oct 2021 18:32 UTC

On 10/26/21 12:23 PM, James Kuyper wrote:
....
> No, I said "it's not permitted to declare functions that take arrays as
> arguments, but it is permitted to declare a function parameter as if it
> were an array."
>
> Such a declaration does NOT declare the parameter to be an array, it
> declares it to be a pointer. You've retained my citation of the part of
> the standard that says so above. Here's a complete compilable program
> demonstrating that rule:
>
> #include <stdio.h>
>
> static void func(int a[2][3])
> {
> printf("%s\n",
> _Generic(a, int[2][3]: "array", int(*)[3]: "pointer"));
> }
>
> int main(void)
> {
> int a[2][3];
> func(a);
> }
>
> When I run that program, it says "pointer". What do you get when you run it?

Unfortunately, that doesn't prove what I intended it to prove, because
even if a were an array, it would have been converted to a pointer when
passed to _Generic. That problem is not too difficult to work around, I
just have to use the '&' operator:

#include <stdio.h>

#define ARR_PTR(x) _Generic(&x, \
int(*)[2][3]: "array", \
int(**)[3]: "pointer", \
default: "other")

static void func(int a[2][3])
{
int b[2][3];
int (*c)[3];
printf("a:%s\n", ARR_PTR(a));
printf("b:%s\n", ARR_PTR(b));
printf("c:%s\n", ARR_PTR(c));
}

int main(void)
{
int a[2][3];
func(a);
}

I get the following output:
a:pointer
b:array
c:pointer

What do you get?

Subject	Author
Re: I think references should have been const by default	James Kuyper