Junio C Hamano wrote:
> Thomas Rast <trast@xxxxxxxxxxxxxxx> writes:
>
> > * The word regex matches anything that is !isspace().
> >
> > * The word regex does not match '\n'. (This case is not very harmful,
> > but we used to silently cut off at the '\n' which may go against
> > user expectations.)
>
> How expensive to run this check twice, every time word_regex finds a
> match?
It runs the first bullet point for every non-match, and the second
bullet point for every match. So it looks at every input character
exactly once.
> As this is about making sure that we got a sane regex from the user (or a
> builtin pattern), I wonder if we can make it not depend on the payload we
> are matching the regex against. Then before using a word_regex that we
> have not checked, we check if that regex is sane, mark it checked, and do
> not have to do the check over and over again.
Algorithmically it should be easy once you have the finite state
automaton corresponding to the regex: just verify that for every
possible non-terminal state, there is a transition for every
!isspace() character to a state other than "fail to match" or "match
the empty string".
In the implementation, it might be doable if we switch to compat/regex
on all platforms, since we then have ready access to all internal
structures regcomp() creates, including the DFA.
I'll think about at least using compat/regex for a static check of all
*builtin* patterns, which would be superior to the brute force
approach in 4/4.
--
Thomas Rast
trast@{inf,student}.ethz.ch
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