Solved expr(1) do not honor \+ regex

A

airwan

Hello,

On Freebsd 14.1-Release, I tried expr "Hello 7.1" : '.* \([.0-9]\+\)'
which returns empty string, while \+ replaced by * returns 7.1.

The man page says expressions should be basic regular expression, and I found that \+ is. Is it a bug? A mistakes I did? Thanks for your help.
 
airwan said:
[…] and I found that \+ is. […]
Click to expand...
Where’d you find that? It is Basic Regular Expressions (capitalized as if it was a proper name), not basic regular expressions. re_format(7) explains:​
Obsolete (“basic”) regular expressions differ in several respects. | is an ordinary character and there is no equivalent for its functionality. + and ? are ordinary characters, and their functionality can be expressed using bounds ({1,} or {0,1} respectively).​
Click to expand...
 
expr(1) does not evaluate regexps. It evaluates arithmetic expressions. Why would you think it should honor regexps?
 
cy@ said:
expr(1) does not evaluate regexps. It evaluates arithmetic expressions. Why would you think it should honor regexps?
Click to expand...
Yes it does. Because it is described in the manual:

Code: 
expr1 : expr2
             The ":" operator matches expr1 against expr2, which must be a
             basic regular expression.  The regular expression is anchored to
             the beginning of the string with an implicit "^".

             If the match succeeds and the pattern contains at least one
             regular expression subexpression "\(...\)", the string
             corresponding to "\1" is returned; otherwise the matching
             operator returns the number of characters matched.  If the match
             fails and the pattern contains a regular expression subexpression
             the null string is returned; otherwise 0.
 
schweikh said:
Yes it does. Because it is described in the manual:

Code: 
expr1 : expr2
             The ":" operator matches expr1 against expr2, which must be a
             basic regular expression.  The regular expression is anchored to
             the beginning of the string with an implicit "^".

             If the match succeeds and the pattern contains at least one
             regular expression subexpression "\(...\)", the string
             corresponding to "\1" is returned; otherwise the matching
             operator returns the number of characters matched.  If the match
             fails and the pattern contains a regular expression subexpression
             the null string is returned; otherwise 0.
Click to expand...
This works:

Code: 
slippy$ expr "Hello 7.1" : '..* \([.0-9][.0-9]*\)'  
7.1
slippy$

It doesn't support PCRE.

What you want is gexpr from sysutils/coreutils, which does support PCRE.

Code: 
slippy$ gexpr "Hello 7.1" : '..* \([.0-9][.0-9]*\+\)'
7.1
slippy$
 
Thanks, I can mark the post as solved. I simply replace \+ by {1,} as documented. My goal is to patch a port with minimal modification thus gexpr is not an option. Indeed linux expr seems to use extended form.
 
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