Can you tell me the sequence of events that happen (internally in perl
during parsing) for the following code:
sub fun
{
print @_;
}
fun fun (1), fun 2, fun (3);
I am particularly interested in the comma operator, in the above code,
as I understand it
(1) first the list operations (which have highest precedence because
of parantheses) will execute first printing "13".
(2) then the comma between fun(1) and fun 2 is evaluated, once this
happens, is the list operation ( fun 2, fun(3) ) evaluated or is the
comma between 2 and fun(3) evaluated?
Thanks,
Joel
On Mar 20, 9:21 pm, [EMAIL PROTECTED] (Chas. Owens) wrote:
> On Thu, Mar 20, 2008 at 11:28 AM, Joel <[EMAIL PROTECTED]> wrote:
> > > > my_function "foo", "bar";
>
> > > > my_function is a list operator which has a very low precedence so the
> > > > parentheses are not required.
>
> > > Sometimes, i.e. if the sub is not predeclared, they are required.
>
> > yes this is true, because perl doesn't know that my_function is
> > actually a function call when it doesn't see it predeclared..
>
> > I have another question, why are function calls without parantheses
> > called list operators, is there no difference between the two?
> > why isn't a function call with parantheses called a list operator too?
>
> snip
>
> Not all subroutines are list operators. For instance,
>
> sub square ($) { $_[0] ** 2 }
>
> is a named unary operator.
>
> from perldoc perlop*
> Actually, there aren't really functions in this sense, just
> list operators and unary operators
> behaving as functions because you put parentheses around the arguments.
>
> *http://perldoc.perl.org/perlop.html#Terms-and-List-Operators-(Leftward)
> --
> Chas. Owens
> wonkden.net
> The most important skill a programmer can have is the ability to read.
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