Questions on a Regular Rexpression

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Ioannis Vranos - 25 Feb 2005 21:44 GMT

Given the regular expression:

S"^([a-zA-Z]+|[a-zA-z]+\\s[a-zA-Z]+)$"

1) Isn't the "[a-zA-Z]+|[a-zA-z]+" part redundant? As far as I can
understand it means exactly the same as "[a-zA-Z]+" alone.

2) Isn't the parenthesis grouping redundant?

3) How can we define the parenthesis characters themselves as expected
characters in a match?

Thanks in advance.

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Ioannis Vranos

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Carl Daniel [VC++ MVP] - 25 Feb 2005 23:34 GMT

> Given the regular expression:
>
> S"^([a-zA-Z]+|[a-zA-z]+\\s[a-zA-Z]+)$"
>
> 1) Isn't the "[a-zA-Z]+|[a-zA-z]+" part redundant? As far as I can
> understand it means exactly the same as "[a-zA-Z]+" alone.

No, because of the alternative - it's

[a-zA-Z]+

-or-

[a-zA-z]+\\s[a-zA-Z]+

> 2) Isn't the parenthesis grouping redundant?

Since it's the entire expression, yes. If this expression was embedded
inside a larger regex then no - it defines the limits of the alternative.

> 3) How can we define the parenthesis characters themselves as expected
> characters in a match?

Just escape them: \\(. You shouldn't need to escape the right paren in
most cases - just the left.

-cd

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Ioannis Vranos - 26 Feb 2005 08:36 GMT

>>1) Isn't the "[a-zA-Z]+|[a-zA-z]+" part redundant? As far as I can
>>understand it means exactly the same as "[a-zA-Z]+" alone.
[quoted text clipped - 6 lines]
>
> [a-zA-z]+\\s[a-zA-Z]+

I did not understand what you mean with the above. May you explain with
some details?

>>2) Isn't the parenthesis grouping redundant?
>
[quoted text clipped - 6 lines]
> Just escape them: \\(. You shouldn't need to escape the right paren in
> most cases - just the left.

Ok, thanks for the info.

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Ioannis Vranos

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Carl Daniel [VC++ MVP] - 26 Feb 2005 14:36 GMT

>>> 1) Isn't the "[a-zA-Z]+|[a-zA-z]+" part redundant? As far as I can
>>> understand it means exactly the same as "[a-zA-Z]+" alone.
[quoted text clipped - 9 lines]
> I did not understand what you mean with the above. May you explain
> with some details?

The alternative operation has low precendence - lower than concatenation, so

(bob|joe|sue)

parses as 'bob' or 'joe' or 'sue' not as 'bo'+('b' or 'j')+'o'+('e' or
's')+'ue'

similarly,

[a-zA-Z]+|[a-zA-Z]+\\s+[a-zA-Z]+

parses as

'[a-zA-Z]+' or '[a-zA-Z]+\\s[a-zA-Z]+'

instead of

('[a-zA-Z]+' or '[a-zA-Z]+')\\s+[a-zA-Z]+

does that make sense?

The original expression could be factored, since the alternatives have a
common prefix:

[a-zA-Z]+(\\s+[a-zA-Z]+)?

I would expect a DFA-based regex engine might well do that factoring as a
matter of course when computing the DFA.

-cd

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Ioannis Vranos - 26 Feb 2005 15:25 GMT

> The alternative operation has low precendence - lower than concatenation, so
>
[quoted text clipped - 24 lines]
> I would expect a DFA-based regex engine might well do that factoring as a
> matter of course when computing the DFA.

Thanks for the explanation.

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Ioannis Vranos

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Serge Baltic - 26 Feb 2005 18:14 GMT

IV> S"^([a-zA-Z]+|[a-zA-z]+\\s[a-zA-Z]+)$"

Note that the [A-z] character set listed above (in the second group) includes
non-alphabetic characters.

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Serge

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Ioannis Vranos - 26 Feb 2005 19:20 GMT

> IV> S"^([a-zA-Z]+|[a-zA-z]+\\s[a-zA-Z]+)$"
>
> Note that the [A-z] character set listed above (in the second group)
> includes non-alphabetic characters.

Thanks for the correction, it was just a typo of mine, it was meant to be:

S"^([a-zA-Z]+|[a-zA-Z]+\\s[a-zA-Z]+)$"

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Ioannis Vranos

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Questions on a Regular Rexpression

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