Questions on a Regular Rexpression

[Ioannis Vranos]

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.

--
Ioannis Vranos

[Carl Daniel [VC++ MVP]]

Ioannis Vranos wrote:

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

[Ioannis Vranos]

Carl Daniel [VC++ MVP] wrote:

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]+

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

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.

Ok, thanks for the info.

--
Ioannis Vranos

[Carl Daniel [VC++ MVP]]

Ioannis Vranos wrote:

Carl Daniel [VC++ MVP] wrote:

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]+

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

[Ioannis Vranos]

Carl Daniel [VC++ MVP] wrote:

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.

Thanks for the explanation.

--
Ioannis Vranos

[Serge Baltic]

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.

--
Serge

[Ioannis Vranos]

Serge Baltic wrote:

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]+)$"

--
Ioannis Vranos