The second one requires a word boundary to exist immediately to the left and to the right of the ==. Your string, "abc == 123" has white-space immediately to the left and right of the ==. Consequently, the word boundaries are between 'c' and 'space', and again between 'space' and '1'. There is no boundary between 'space' and '=', nor between '=' and 'space'.

Thus, the second RE cannot match against your target string. To put it a little more clearly (I hope): 

abc[boundary](space)==(space)[boundary]123  <---- What your string loo
+ks like.
abc(space)[boundary]==[boundary](space)123  <---- What you're trying t
+o match.

[download]

That second case is impossible; there can never be a word boundary between a space and an equals sign.

Rather than using non-greedy quantifiers, why don't you instead specify what would be considered legal on the left-hand side of the ==. It's likely that your left-hand side needs to be an identifier of some sort. If that's the case, specify what characters are legal: 

m/^(\w+)\s*==\s*(\w+)$/

[download]

If either side can contain quoted literals that may embed '==', you're going to have to use a real parser instead.


Dave

In reply to Re: Regular Expression Question by davido
in thread Regular Expression Question by VingInMedina

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