Linda Walsh <[EMAIL PROTECTED]> wrote:
> The longest matching substring (because you use "/" to start the pattern,
> yes?)
No, !() normally uses the longest match, just like *, *(), and +().
> s="thomas rich george"
>
> Manpage says "!()" will match anything except one of the patterns.
>
> "$s" is 1 pattern, no? (no alternation); 1 pattern containing
> "thomas rich george", yes?
Right.
> Then if I use !($s) as the replacement pattern within in the substitute
> operator, ( ${s//!($s)/X} ) the !($s) should match any string
> except what is in "$s" ("thomas rich george").
It will match the longest substring of $s that is not $s itself -
which is the first N-1 characters of s. With //, it will replace
every match, so the final character will also be replaced, since it is
not the same as all of $s.
> So in the substitute, the string to replace (the "!($s) part) should
> match nothing in my variable "$s", so I'd think no replacement would
> be done.
You're right that $s as a whole does not match !($s). But that isn't
the only candidate for matching. Every substring is a candidate. The
first position where a match is found is used, with the longest match
starting at that position.
To check whether a string as a whole matches a pattern, instead of
checking for a matching substring, you can use case. (Or, if the
pattern is a literal string with no special characters, you can use
test.)
>> echo \"${s//!($s)/X}\"
> "XX" # why two X's? if I use 1 "/" instead of double:
>> echo \"${s/!($s)/X}\"
> "Xe" # why an "e" afterwards?
With / instead of //, only the first match is replaced.
paul
