Re: Command Languages Versus Programming Languages

In article <87cyinrt5s.fsf@doppelsaurus.mobileactivedefense.com>,
Rainer Weikusat  <rweikusat@talktalk.net> wrote:

scott@slp53.sl.home (Scott Lurndal) writes:

Rainer Weikusat <rweikusat@talktalk.net> writes:

>
[...]
>

Something which would match [0-9]+ in its first argument (if any) would
be:
>
#include "string.h"
#include "stdlib.h"
>
int main(int argc, char **argv)
{
   char *p;
   unsigned c;
>
   p = argv[1];
   if (!p) exit(1);
   while (c = *p, c && c - '0' > 10) ++p;
   if (!c) exit(1);
   return 0;
}
>
but that's 14 lines of text, 13 of which have absolutely no relation to
the problem of recognizing a digit.

>
Personally, I'd use:
>
$ cat /tmp/a.c
#include <stdint.h>
#include <string.h>
>
int
main(int argc, const char **argv)
{
    char *cp;
    uint64_t value;
>
    if (argc < 2) return 1;
>
    value = strtoull(argv[1], &cp, 10);
    if ((cp == argv[1])
     || (*cp != '\0')) {
        return 1;
    }
   return 0;
}

>
This will accept a string of digits whose numerical value is <=
ULLONG_MAX, ie, it's basically ^[0-9]+$ with unobvious length and
content limits.

He acknowledged this already.

return !strstr(argv[1], "0123456789");
>
would be a better approximation,

No it wouldn't.  That's not even close.  `strstr` looks for an
instance of its second argument in its first, not an instance of
any character in it's second argument in its first.  Perhaps you
meant something with `strspn` or similar.  E.g.,

    const char *p = argv[1] + strspn(argv[1], "0123456789");
    return *p != '\0';

just a much more complicated algorithm
than necessary. Even in strictly conforming ISO-C "digitness" of a
character can be determined by a simple calculation instead of some kind
of search loop.

Yes, one can do that, but why bother?

- Dan C.