Don't you mean
/\Q$foo\E/ and print "$&\n";
s/\Q$foo\E/XYZ/;
Using -w would have helped you figure it out.
Update: oh right, deliberate tyops. Well,
now that I've really thought about it...
this is what is going on. The line
/\Q$fooE/ and print "$&\n";
given that $fooE is not defined is the same thing as
/\Q/ and print "$&\n";
which is the same thing as
// and print "$&\n";
Which is the same thing as finding the same thing as the
last successful match. This also applies to the s expression,
which is why your bar appears to be magically transmuting
itself into XYZ.
That's not a bug, that's a feature.
--
g r i n d e r
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