How does this "JAPH" work?

[kcorj2244]

I've been stumped on this one for a while:

$_ = "wftedskaebjgdpjgidbsmnjgc";
tr/a-z/oh, turtleneck Phrase Jar!/; print;

Prints : "Just another Perl Hacker,"

[talexb]

$_ = "wftedskaebjgdpjgidbsmnjgc";
tr/a-z/oh, turtleneck Phrase Jar!/; print;
       abcdefghijklmnopqrstuvwxyz
            |                |
            |                +- J
            +------------------ u
                                ..

I've expanded the alphabet (a-z) and aligned it with the phrase it's being replaced with, to show you how the first two characters map. The first w maps to the J in Jar, the next f maps to the u in 'turtleneck', and so forth. Simple! :)

[AnomalousMonk]

talexb has shown how this simple substitution works. Note, however, that the output is not "Hacker" as you have it in the OP, but "hacker" (little-h vice big-H). To verify your understanding of what's going on, what simple change to the replacement string of the  tr/// operator would be necessary to make the output "Hacker"? (Of course, this simple change screws up another thing, but that's life... :)

[davido]

There are three statements in this JAPH:

1. $_ = "wftedskaebjgdpjgidbsmnjgc";
2. tr/a-z/oh, turtleneck Phrase Jar!/;
3. print;

Step one, assign some encrypted text to $_; the "it" variable (Perl's default variable). See perlvar for an explanation of $_.

Step two, transliterate. See perlop for an explanation of how tr/abc/def/ works, but the gist is that it walks through a string and changes any chcaracter found in the "abc" group to the character in the same position in the "def" list. Hence, if the string contains a "a", it would be replaced with a "d". The string that it acts upon by default is the one contained in $_.

Step three, print the contents of $_. print prints the contents of $_ if there is no other argument passed to it.

This is essentially a form of substitution cipher that leverages a couple of the terse shortcuts Perl offers to achieve a small degree of code illegibility.