Re: how to translate lua pattern "(.*)and(.*)" with lpeg re ?
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- Subject: Re: how to translate lua pattern "(.*)and(.*)" with lpeg re ?
- From: Sean Conner <sean@...>
- Date: 2018年1月20日 17:54:27 -0500
It was thus said that the Great albertmcchan once stated:
> local C, P = lpeg.C, lpeg.P
>
> -- my attempt for lua pettern "(.+)and(.*)"
> local lpeg_pat = C((P(1) - 'and')^1) * 'and' * C(P(1)^0)
> local re_pat = re.compile "{ (. ! 'and')* . } 'and' {.*}"
>
> -- lua pattern "(.*)and(.*)
> lpeg_pat = C((P(1) - 'and')^0) 'and' * C(P(1)^0)
>
> -- what is lpeg re equivalent code ?
That was weird. I would expect
C((P(1) - 'and')^1) * 'and' * C(P(1)^0)
to map to:
{ ( . ! 'and' )+ } 'and' { .* }
but it didn't. I went so far as to recompile LPeg with debugging so I could
dump the parse tree. The LPeg that works dumped out:
[]
seq
seq
capture kind: 'simple' key: 0
seq
seq
not
seq
char 'a'
seq
char 'n'
char 'd'
any
rep
seq
not
seq
char 'a'
seq
char 'n'
char 'd'
any
seq
char 'a'
seq
char 'n'
char 'd'
capture kind: 'simple' key: 0
rep
any
While the re code dumped out as:
[]
seq
seq
capture kind: 'simple' key: 0
seq
seq
any -- !!!!!!
not
seq
char 'a'
seq
char 'n'
char 'd'
rep
seq
any
not
seq
char 'a'
seq
char 'n'
char 'd'
seq
char 'a'
seq
char 'n'
char 'd'
capture kind: 'simple' key: 0
rep
any
The difference is marked. I then tried:
{ (! 'and' . )+ } 'and' {.*}
(NOTE: I swapped the order in the first bit) and that produced the same
code as the LPeg version. So to answer your question, the re equivalent
code would be:
{ (! 'and' .)* } 'and' {.*}
-spc