Re: Top 10 most common hard skills listed on resumes...

David Brown <david.brown@hesbynett.no> writes:

On 09/09/2024 20:46, Kaz Kylheku wrote:

On 2024-09-09, David Brown <david.brown@hesbynett.no> wrote:

On 09/09/2024 18:57, Bart wrote:

On 09/09/2024 17:21, Waldek Hebisch wrote:

Keith Thompson <Keith.S.Thompson+u@gmail.com> wrote:

>

C23 doesn't add any new support for 128-bit integers.

>
So what does _Bitint do with a width of 128 bits?
>

>
_BitInt types are not "integer types".  Nor is gcc's __int128 type.

How can we write a program which, in an implementation which has a
__int128 type, outputs "yes" if it is an integer type, otherwise "no"?

I'm not sure such a program, one that can detect either an __int128 that
isn't an integer type or an __int128 that is an integer type, is possible.

#include <stdio.h>
>
int main() {
    auto const x = 0x1'0000'0000'0000'0000;
    if (x > 0xffff'ffff'ffff'ffff) {
        printf("yes\n");
    } else {
        printf("no\n");
    }
}

Of course this uses digit separators, which are a new feature in C23.

If an implementation doesn't have an integer type wider than 64 bits,
then the constant 0x1'0000'0000'0000'0000 has no type, which violates a
constraint.

If an implementation does have a integer types wider than 64 bits,
there's no guarantee that it uses the name "__int128".  A future gcc
might have a 128-bit (extended) integer type and still have __int128
with its current semantics.

For gcc, this meets Kaz's specification:

#include <stdio.h>
int main(void) {
    puts("no");
}

--
Keith Thompson (The_Other_Keith) Keith.S.Thompson+u@gmail.com
void Void(void) { Void(); } /* The recursive call of the void */