Re: is expr's \? regex metachar working?

[Andreas Schwab]

"Clark O. Morgan" <address@hidden> writes:

|> Forgive me if this is a dup post.  A previously posted copy of this article
|> hasn't shown up on aracnet's news server or on newsguy....
|> 
|> 
|> The info page for expr 2.0 says that with respect to the ':' operator's
|> regular expression argument:
|> 
|>      In the regular expression, `\+', `\?', and `\|' are operators
|>      which respectively match one or more, zero or one, or separate
|>      alternatives.
|> 
|> I can't get `\?' to work--it always seems to return
|> the length of the target string.....
|> 
|> $ ./expr --version
|> expr (GNU sh-utils) 2.0
|> Written by Mike Parker.
|> 
|> $ ./expr "abcdef" : '.*b'
|> 2
|> $ ./expr "abcdef" : '.*u'
|> 0
|> $ ./expr "abcdef" : '.*u\?'
|> 6
|> $ ./expr "abcdef" : '.*b\?'
|> 6                           <--- this looks wrong

No, this is correct.  The '.*' regexp matches the maximum possible length,
which is all 6 characters.  Since b is optional, it does not cause a match
failure.

|> $ ./expr "abcdef" : '.*z\?'
|> 6
|> $ ./expr "abcdef" : '.*\(z\)\?'
|> 
|> $ ./expr "abcdef" : '.*\(b\)\?'
|>                            <--- this looks wrong

Same here.  \(b\) is not used for matching, thus \1 is empty.

Andreas.

-- 
Andreas Schwab                                  "And now for something
SuSE Labs                                        completely different."
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