Thank you for the reply. I'm obviously not understanding something as I thought my (?{...}) block was at the very end of the regular expression. I didn't think it would be encountered and triggered unless both look-behind and look-ahead succeeded. I expected the matching to go something like this

Initialise pointer to beginning of string
test look-behind, nothing preceding so fails
advance pointer one place
test look-behind, '{' so fails
advance pointer one place
test look-behind, 'x' so fails
advance pointer one place
test look-behind, '1' so fails
advance pointer one place
test look-behind, '}' so succeeds
test look-ahead, '[' also succeeds
code block encountered, execute
advance pointer one place
test look-behind, '[' so fails
...

[download]

If the DFA had to look past my code block to check something else and then backtrack, the behaviour would make sense. However, I can't see how that is happening in this case.

I'm sure it must be me failing to get my head around something fundamental. Please could you point out where my understanding is lacking.

Cheers,

JohnGG

In reply to Re^2: Regex code block executes twice per match using look-arounds by johngg
in thread Regex code block executes twice per match using look-arounds by johngg

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