Re^2: perl5 "prototyping": why?

Thanks, I was thinking the same thing. But what if the arguments are complex?

sub foo (\$\@\%) {
  my ( $bar, @baz, %something ) = @_;
}
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...erm, I'm just making up that example (obviously), and I never bother with prototypes so this may well be all wrong, but wouldn't @baz suck in the remainder of @_ after $bar if you remove the prototype? Or something like that?

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Re^3: perl5 "prototyping": why? by CountZero (Bishop) on Jul 07, 2005 at 21:19 UTC
Perldoc says: Any backslashed prototype character represents an actual argument that absolutely must start with that character. The value passed as part of @_ will be a reference to the actual argument given in the subroutine call, obtained by applying \ to that argument. So you call `foo` as `foo($a_scalar, @an_array, %a_hash)` and you will get automatic reference/dereference magic. Dropping the backslash however breaks this behaviour: Unbackslashed prototype characters have special meanings. Any unbackslashed @ or % eats all remaining arguments, and forces list context. An argument represented by $ forces scalar context. `@baz` would then indeed gobble up the hash following it. CountZero "If you have four groups working on a compiler, you'll get a 4-pass compiler." - Conway's Law	[reply] [d/l] [select]

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Re^3: perl5 "prototyping": why?
by CountZero (Bishop) on Jul 07, 2005 at 21:19 UTC

Any backslashed prototype character represents an actual argument that absolutely must start with that character. The value passed as part of @_ will be a reference to the actual argument given in the subroutine call, obtained by applying \ to that argument.

foo

foo($a_scalar, @an_array, %a_hash)

Dropping the backslash however breaks this behaviour:

Unbackslashed prototype characters have special meanings. Any unbackslashed @ or % eats all remaining arguments, and forces list context. An argument represented by $ forces scalar context.

@baz would then indeed gobble up the hash following it.

CountZero

"If you have four groups working on a compiler, you'll get a 4-pass compiler." - Conway's Law

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