note
fullermd
<p>
Interesting. It does sound buggy.
</p>
<p>
And it does the same thing with brackets too, notice. Simpler example:
</p>
<code>
@s = ("foo bar baz") x 4;
$bar = "rab";
%bar = (1 => "eek");
@bar = (undef, 'oof');
$s[0] =~ s{(bar)} {$bar\{1\}};
$s[1] =~ s{(bar)} [$bar\{1\}];
$s[2] =~ s{(bar)} {$bar\[1\]};
$s[3] =~ s{(bar)} [$bar\[1\]];
for $i (0..3)
{ print "$i -> '$s[$i]'\n"; }
</code>
<p>Yields</h2>
<code>
0 -> 'foo eek baz'
1 -> 'foo rab{1} baz'
2 -> 'foo rab[1] baz'
3 -> 'foo oof baz'
</code>
<p>(5.8 here)</p>
<p>
So something about having the delimiter be the same as your subscripting character makes it real hard to escape.
</p>
<p>
For extra fun, notice that using two \'s in the sub fails as "replacement not terminated", and using 3 leaves one sitting in your string:
</p>
<code>
4 -> 'foo rab\[1] baz'
</code>
<p>
So it doesn't seem like you can all that easily work around it except by changing the delimiters. Maybe playing with escaping... yes, that seems to work:
</p>
<code>
$s[4] =~ s{(bar)} [$bar\E\[1\]];
4 -> 'foo rab[1] baz'
</code>
<p>
Wackiness.
</p>
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