Ken Irving wrote:
>> $ printf "%d %s\n" 1 ok -
>> 1 ok
>> -bash: printf: -: invalid number
>> 0
>>
>> why getting error here, and not in the previous?
>> why "invalid number" ?
>> what is that zero?
>
> Again, you have more arguments than operators, so it makes another pass,
> and on the second pass tries to format - as a number... Don't know
> about the zero, but I guess %d maybe starts with a default of 0...?
When bash detects an integer conversion error, it returns 0. That's not
strictly Posix-conformat, so the next version of bash will return
whatever strtoimax/strtoll/strtol returns, which is supposed to be the
value accumulated up to the point of the error.
Chet
--
``The lyf so short, the craft so long to lerne.'' - Chaucer
``Ars longa, vita brevis'' - Hippocrates
Chet Ramey, ITS, CWRU c...@case.edu http://cnswww.cns.cwru.edu/~chet/
