The underscore prototype was introduced in Perl 5.10. In the 2008 node Re: Underscore _ prototype in 5.10, lodin explains it as follows:

The effect of _ is that if the argument is missing, $_ is passed, and thus populates @_.

See the example code there, and also see this blog post.

Hope that helps,

From perlsub:

As the last character of a prototype, or just before a semicolon, a @ or a % , you can use _ in place of $ : if this argument is not provided, $_ will be used instead.

So I guess _ being the only character in the prototype, is also the last, and is equivalent to $ there.

Update: I see from the response of Athanasius above that my guess is slightly off. I didn't quite understand what the documentation is saying.

If you are referring to my node, there's probably nothing wrong with the documentation, just my quick reading of it. When it says you can use _ in place of $ I got the idea that prototypes (_) and ($) would be the same, which is not the case.

sub foo(_){ warn @_ }
sub goo($){ warn @_ }
foo();
goo();
[download]

output:
Not enough arguments for main::goo at pm3 line 4, near "()"
Execution of pm3 aborted due to compilation errors.

Whats wrong with the documentation?

It didn't show up in the search :)

It doesn't mention  (_) and  (_) is not on the list Declared as            Called as

TIP: a file-search in perlsub#Prototypes for  _ (surrounded by spaces) helps! =)

Cheers Rolf

( addicted to the Perl Programming Language)