How to properly look with LPEG
To match a line starting with dog
and then cat
(but not consuming cat
) this works:
local lpeg = require 'lpeg'
local str1 = 'dogcat'
local patt1 = lpeg.C(lpeg.P('dog')) * #lpeg.P('cat')
print(lpeg.match(patt1, str1))
Output: dog
To match a line starting with dog
, it is followed by any sequences of characters, and then cat
(but does not consume it), as a regex lookup (dog.+?)(?=cat)
, I tried this:
local str2 = 'dog and cat'
local patt2 = lpeg.C(lpeg.P("dog") * lpeg.P(1) ^ 1) * #lpeg.P("cat")
print(lpeg.match(patt2, str2))
My expected result dog and
, but it returns nil
.
If I remove part of the overview (i.e. using a pattern lpeg.C(lpeg.P("dog") * lpeg.P(1) ^ 1)
) it can match the whole line successfully. That means the part * lpeg.P(1) ^ 1
correctly matches any sequence of characters, doesn't it?
How to fix it?
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You need to undo "cat" at every position in the lookahead that might match:
local patt2 = lpeg.C(lpeg.P"dog" * (lpeg.P(1)-lpeg.P"cat") ^ 1) * #lpeg.P"cat"
I find it advisable to include the debugger I was working on ( pegdebug ) as it helps in cases like this. Here is the output it generates for the original lpeg expression:
+ Exp 1 "d"
+ Dog 1 "d"
= Dog 1-3 "dog"
+ Separator 4 " "
= Separator 4-11 " and cat"
+ Cat 12 ""
- Cat 12
- Exp 1
You can see that the Separator expression "eats" all characters, including "cat", and there is nothing to match with P"cat"
.
The result for the modified expression looks like this:
+ Exp 1 "d"
+ Dog 1 "d"
= Dog 1-3 "dog"
+ Separator 4 " "
= Separator 4-8 " and "
+ Cat 9 "c"
= Cat 9-11 "cat"
= Exp 1-8 "dog and "
/ Dog 1 0
/ Separator 4 0
/ Exp 1 1 "dog and "
Here's the complete script:
require 'lpeg'
local peg = require 'pegdebug'
local str2 = 'dog and cat'
local patt2 = lpeg.P(peg.trace { "Exp";
Exp = lpeg.C(lpeg.V"Dog" * lpeg.V"Separator") * #lpeg.V"Cat";
Cat = lpeg.P("cat");
Dog = lpeg.P("dog");
Separator = (lpeg.P(1) - lpeg.P("cat"))^1;
})
print(lpeg.match(patt2, str2))
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