Dynamic execution. Is there such a thing? 3

[ejaggers]

Can you pass dynamically execute code in Perl. i.e.

chomp($subname = <STDIN>);
$y = $subname();

 

[feherke]

Hi

Take a look at thread219-1496647 , it may answer your question.

Feherke.

http://rootshell.be/~feherke/

 

[prex1]

[tt]eval[/tt] is there for that purpose.

Franco

http://www.xcalcs.com

: Online engineering calculations

http://www.megamag.it

: Magnetic brakes for fun rides

http://www.levitans.com

: Air bearing pads

 

[KevinADC]

What he appears to want to do is use a hash where the subroutine name is the hash key and the value is a reference to code or a subroutine. He can then use the name of the subroutine as a reference to the subroutine.

------------------------------------------
- Kevin, perl coder unexceptional!

 

[ejaggers]

Kevin,

Actually what I have is: sub1(),sub2()...sub20();

Is there a way to say:

chomp($x = <STDIN> );
sub$x();

 

[feherke]

Hi

As Kevin already mentioned, this :

sub sub1 { print "one\n" }

sub sub2 { print "two\n" }

sub sub3 { print "three\n" }

foreach $x (1..3) {
  print "running sub #$x :\n";
  &{"sub$x"};
}

will output this :
[tt]
running sub #1 :
one
running sub #2 :
two
running sub #3 :
three
[/tt]
Now go ahead and try it out.

Feherke.

http://rootshell.be/~feherke/

 

[ejaggers]

Kevin,

please give an example of how to do it your way
I'm not that familiar with references

 

[feherke]

Hi

I'm not that familiar with references

http://perldoc.perl.org/perlref.html

Feherke.

http://rootshell.be/~feherke/

 

[ejaggers]

I tried your solution above, and went from 67 lines of code down to 26. That was awsome!!! Thanks again

And thanks for the link to references.

 

[feherke]

Hi

Actually I just learned this from Kevin at 18 Dec 08 3:01. So I pass the thanks to him.

Feherke.

http://rootshell.be/~feherke/

 

[KevinADC]

Using a hash example:

use strict;
use warnings;

my %coms = (
   foo => \&foo,
   bar => \&bar,
);

foreach my $x qw(foo bar) {
  print "running sub $x :\n";
  $coms{$x}->($x);
}

sub foo {
   my $arg = $_[0];
   print "in $arg sub\n";
}

sub bar {
   my $arg = $_[0];
   print "in $arg sub\n";
}

------------------------------------------
- Kevin, perl coder unexceptional!