Besides the "secret eskimo greeting" and the "goatse" operator, which other "secret" operators have you heard of?

Secret eskimo greeting: }{

Goatse operator: =()=

Moved to Meditations by castaway

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Re: "Secret" operators
by hardburn (Abbot) on Jan 28, 2005 at 17:48 UTC

You may be familer with the spaceship operator already (<=>). What about the Super Star Destroyer operator?

```\${-} <=\$vader=> \${-}

Update: Now complete with TIE Intercepters!

"There is no shame in being self-taught, only in not trying to learn in the first place." -- Atrus, Myst: The Book of D'ni.

Ok, I tried it out and now I have some doubts...

\$a <=\$b=> \$c is equal to ( \$a <= \$b , \$c), that one is easy...

But what happens if I just use <=\$something=>? Because printing that prints an equal sign, the contents of \$something and another equal sign... I'm missing something, here...

That's the <> operator. =\$something= isn't a simple scalar, so it's equivalent to glob("=\$something="). On the other hand, <\$something> means readline(\$something). (For special weirdness, <\$ something> is the same as glob(\$something).)

See glob, readline and I/O Operators in perlop for details.

I tried <=\$vader=> (and also <==>), but they are both syntax errors. That's why I added the TIEs :)

"There is no shame in being self-taught, only in not trying to learn in the first place." -- Atrus, Myst: The Book of D'ni.

Re: "Secret" operators
by TedYoung (Deacon) on Jan 29, 2005 at 04:30 UTC

So why is it called the Goatse operator?

Ted Young

(\$\$<<\$\$=>\$\$<=>\$\$<=\$\$>>\$\$) always returns 1. :-)

If you don't already know, then you really don't want to know. I really, really, really wish that I didn't know.

Does it have something to do with that certain picture on that certain website... oh. Never mind.

Ted Young

(\$\$<<\$\$=>\$\$<=>\$\$<=\$\$>>\$\$) always returns 1. :-)
Do tell, do tell :-)

I know what it does, but I have no idea why it's named "goatse" (the eskimo thing is trivial, but this one doesn't seem to be).

I heard it had been named after another operator... but in that case, which operator is that, and how come the name?

Re: "Secret" operators
by eyepopslikeamosquito (Chancellor) on Mar 06, 2005 at 08:26 UTC
Actually, *I* started that thread :-)
Re: "Secret" operators
by ambrus (Abbot) on Oct 20, 2014 at 10:38 UTC