matching !(patterns)

[Linda Walsh]

Paul Jarc wrote:
> It looks for substrings in both cases - specifically, the longest
> matching substring, which might happen to be the entire string.  With
> !(), that is often the case.
>
> >     x=12ab34; echo ${x//+([0-9])/X}        # prints XabX
> >     x=12ab34; echo ${x//!(+([0-9]))/X}     # prints X
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The longest matching substring (because you use "/" to start the pattern,
yes?)  I'm not sure I fully understand this behavior though. Let's have
   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?

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").

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.  I.e. the output would be unchanged.  But when I
try it that's not what I get:

> 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?
-----

The converse works as expected:
> echo \"${s//+($s)/X}\" 
"X"
> echo \"${s/+($s)/X}\" 
"X"