Don't you mean

  /\Q$foo\E/ and print "$&\n";

  s/\Q$foo\E/XYZ/;
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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";
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given that $fooE is not defined is the same thing as

/\Q/ and print "$&\n";
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which is the same thing as

// and print "$&\n";
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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.

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g r i n d e r