can this perl code be simplified?

Question

foreach my $f($query->param) {
    foreach my $v($query->param($f)) {
        switch($f) {
            case 'a' {
                switch($v) {
                    case 1 { code1; }
                    case 2 { code2; }
                    case 3 { code3; }
                }

            case 'b' {
                switch($v) {
                    case 1 { code4; }
                    case 2 { code5; }
                }

            case 'c' {
                switch($v) {
                    case 1 { code6; }
                    case 2 { code7; }
                    case 3 { code8; }
                    case 4 { code9; }
                }
            }
        }
    }
}

Looks pretty simple already. And it's hard to tell, since you are not showing what happens in the "code" section. — TLP, Jul 31 '11 at 19:38
Every time you write `code;` does it denote a unique code block? If not then please update your question to show which code blocks are unique and which are duplicates. — Dave, Jul 31 '11 at 19:41
A quick sidenote: as of Perl 5.12, the switch statement [has been deprecated](http://stackoverflow.com/questions/2630547/why-is-the-switch-module-deprecated-in-perl), and been replaced by [`given/when`](http://perldoc.perl.org/perlsyn.html#Switch-statements). — Mike, Jul 31 '11 at 19:53
given/when were in 5.10, and you really should use them, not Switch.pm — ysth, Jul 31 '11 at 19:58

mikegrb · Accepted Answer · 2011-07-31T20:14:54.817

12

Above all... DO NOT USE Switch.pm. If you are using Perl 5.10 or newer, given/when is a native switch statement.

A dispatch table is the solution you seek. http://www.perlmonks.org/?node_id=587072 describes dispatch tables, a technique for executing code via a hash based on some value matching the hash key.

Edit, example:

my %dispatch_table = (
    'a' => {
        '1' => sub { code1; },
        '2' => sub { code2; },
        '3' => sub { code3; },
    'b' => {
        '1' => \&some_sub,
        '2' => sub { code4; },
    }
)

if ( exists( $dispatch_table{$f} ) and exists( $dispatch_table{$f}{$v} ) ) {
    $dispatch_table{$f}{$v}->();
}
else {
    # some default
}

edited Jul 31 '11 at 20:14

answered Jul 31 '11 at 19:50

mikegrb

1,183
5
13

I was going to suggest dispatch tables, too. – Dallaylaen Jul 31 '11 at 19:52
1

@ysth, you don't "flush out" an answer, you "flesh out" an answer :-) http://www.urbandictionary.com/define.php?term=flush%20out – tadmc Jul 31 '11 at 23:02
@tadmc, if it's a really _bad_ answer, you might. – cjm Aug 01 '11 at 03:26

score 3 · Answer 2 · answered Jul 31 '11 at 20:27

3

Put it in a "dispatch table":

my %code = (
    a => {
        1 => sub { code1 },
        2 => sub { code2 },
        3 => sub { code3 },
    },
    b => {
        1 => sub { code4 },
        2 => sub { code5 },
    },
    c => {
        1 => sub { code6 },
        2 => sub { code7 },
        3 => sub { code8 },
        4 => sub { code9 },
    },
);

Then once you have $f and $v, call the correct subroutine:

$code{$f}{$v}->();

answered Jul 31 '11 at 20:27

tadmc

3,714
16
14

sorry mikegrb, I confused it with your answer when I came back to vote – Lem0n Aug 02 '11 at 17:05

score 1 · Answer 3 · answered Jul 31 '11 at 19:42

1

Depending on what you mean by "simplified", you might consider something like this (if you're sure that neither $f nor $v can contain a ','):

foreach my $f ($query->param) {
    foreach my $v ($query->param($f)) {
        switch ("$f,$v") {
            case "a,1" { code; }
            case "a,2" { code; }
            case "a,3" { code; }

            case "b,1" { code; }
            case "b,2" { code; }

            case "c,1" { code; }
            case "c,2" { code; }
            case "c,3" { code; }
            case "c,4" { code; }
        }
    }   
}

(I'm assuming that all the occurrences of code; are actually different.)

answered Jul 31 '11 at 19:42

Keith Thompson

254,901
44
429
631

Why the restriction on the comma? If there was a $f such that $f equals "x,y" and there was a case "x,y,1" and "x,y,2" to address it, this would still work; wouldn't it? – Dave Jul 31 '11 at 19:48
2

@Dave: because you couldn't differentiate `$f="x,y";$v="z"` from `$f="x";$v="y,z"` – ysth Jul 31 '11 at 19:59

can this perl code be simplified?

3 Answers3