I was looking for something in perlop when I noticed this in the description for the tr/// operator:

 If no string is specified via the =~ or !~ operator, the $_ string is
+ transliterated.
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Never having thought of using !~ in conjunction with tr///, I wondered what it did, so I tried a few things:

C:\test>perl -wle"$_ = 'aaa'; print $_ !~ tr[a][]"

C:\test>perl -wle"$_ = 'aaa'; print $_ !~ tr[b][]"
1
C:\test>perl -wle"$_ = 'aaa'; print $_ !~ tr[bb][]"
1
C:\test>perl -wle"$_ = 'aaa'; print $_ !~ tr[bc][]"
1
C:\test>perl -wle"$_ = 'aaa'; print $_ !~ tr[b][]c"

C:\test>perl -wle"$_ = 'aaa'; print $_ !~ tr[b][]d"
1
C:\test>perl -wle"$_ = 'aaa'; print $_ !~ tr[b][]"
1
C:\test>perl -wle"$_ = 'aaa'; print $_ !~ tr[a][]c"
1
C:\test>perl -wle"$_ = 'aaa'; print $_ !~ tr[ab][]"

C:\test>perl -wle"$_ = 'aaa'; print $_ !~ tr[a][]"

C:\test>perl -wle"$_ = 'aaa'; print $_ !~ tr[b][]"
1
C:\test>perl -wle"$_ = 'aaa'; print $_ !~ tr[c][]"
1
C:\test>perl -wle"$_ = 'aaa'; print $_ !~ tr[bc][]"
1
C:\test>perl -wle"$_ = 'aab'; print $_ !~ tr[bc][]"
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From which I concluded that in this 'inverted count-the-stars' mode, it returns a true or false value indicating whether the string doesn't contain any of the characters in the searchlist. Not exactly intuative, but could be useful sometime. So then I thought I'd try the delete flag, and the result surprised me no end.

Can you guess what this would do before you try it?

perl -wle"$_ = 'aaa'; $_ !~ tr[a][]d; print"
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Were you right?

Can you explain it ;)