On Nov 5, Tore Aursand said:
>On Tue, 04 Nov 2003 13:58:41 -0500, mgoland wrote:
>> %hash=( 1 => funca(),
>> 2 => funcb(),
>> );
>
>Try this:
>
> %hash = (1 => \&funca(),
> 2 => \&funcb());
No; \&foo is a reference to the foo function, but \&foo() is a reference
to the re
On Tue, 04 Nov 2003 13:58:41 -0500, mgoland wrote:
> %hash=( 1 => funca(),
> 2 => funcb(),
> );
Try this:
%hash = (1 => \&funca(),
2 => \&funcb());
--
Tore Aursand <[EMAIL PROTECTED]>
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Why do the subs execute? Because you told them to. When you type
funca(), you're telling Perl to execute funca, but don't pass it any
parameters. You don't "declare" subs in Perl the way you would in C.
The only place you declare a sub is in the "sub mySub{BLOCK}" statement.
You CAN prototype a
Mark wrote:
>
> I am trying to create a hash where each key is a referance to
> subs. The bellow code produces the expected result, with one
> unexpected side-effect. Why are the subs executed when I only
> declare them ??
>
> <- cut
> #!/usr/local/bin/perl -w
>
>
> %hash=( 1 => funca(),
>
On Tuesday, November 4, 2003, at 12:58 PM, [EMAIL PROTECTED] wrote:
Hello Perler's
I am trying to create a hash where each key is a referance to subs.
The bellow code produces the expected result, with one unexpected
side-effect. Why are the subs executed when I only declare them ??
<- cut
#!
