benjamin gerard <[EMAIL PROTECTED]> writes:
> $expr 'a0b' : 'a\(.\)b' \| '1
>
> The previous sample returns '1'. I assume this is because the matching
> pattern is '0' and it is evaluate as false.
>
> I am just wandering if this is expected behaviour ?
You have just answered you question above.
expr1 | expr2 Returns the evaluation of expr1 if it is neither null nor
zero; otherwise, returns the evaluation of expr2 if it is
not null; otherwise, zero.
Andreas.
--
Andreas Schwab, SuSE Labs, [EMAIL PROTECTED]
SuSE Linux AG, Maxfeldstra�e 5, 90409 N�rnberg, Germany
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"And now for something completely different."
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