On Fri, Mar 11 2022, Eric Sunshine wrote:
> On Wed, Mar 9, 2022 at 3:14 PM Ævar Arnfjörð Bjarmason <avarab@xxxxxxxxx> wrote:
>> On Wed, Mar 09 2022, Junio C Hamano wrote:
>> > Eric Sunshine <sunshine@xxxxxxxxxxxxxx> writes:
>> >> This seems to work, though it's getting a bit verbose:
>> >>
>> >> awk '/^glibc / { split($2,v,"."); if (sprintf("%s.%s", v[1], v[2])
>> >> - 2.34 < 0) exit 1 }'
>> >
>> > In general it is a good discipline to question a pipeline that
>> > preprocesses input fed to a script written in a language with full
>> > programming power like awk and perl (and to lessor extent, sed) to
>> > see if we can come up with a simpler solution without pipeline
>> > helping to solve what these languages are invented to solve, and I
>> > very much appreciate your exploration ;-)
>>
>> I agree :) But the first language we've got here is C. Rather than
>> fiddle around with getconf, awk/sed etc. why not just the rather
>> trivial:
>>
>> +#include "test-tool.h"
>> +#include "cache.h"
>> +
>> +int cmd__glibc_config(int argc, const char **argv)
>> +{
>> +#ifdef __GNU_LIBRARY__
>> + printf("%d\n%d\n", __GLIBC__, __GLIBC_MINOR__);
>> + return 0;
>> +#else
>> + return 1;
>> +#endif
>> +}
>
> It feels overkill to add this just for this one case which is
> otherwise done easily enough with existing shell tools.
>
> That said, perhaps I'm missing something, but I don't see how this
> solution helps us get away from the need for `expr`, `awk`, or `perl`
> since one of those languages would be needed to perform the arithmetic
> comparison (checking if glibc is >= 2.34).
In this case they're \n delimited, so we can use the shell's native
whitespace splitting to $(())-compare $1 and $2.
But probably better is to just amend that to call it as "test-tool libc
is-glibc-2.34-or-newer" or whatever. Then just do:
if (__GLIBC__ > 2 || (__GLIBC__ == 2 && 34 >= __GLIBC_MINOR__))
return 0;
return 1;
