Re: Passing by Named Parameter

I might do something like this:

sub mysub {
    %params = %{ +shift };
    my %known_options = ( 'this' => 1, 'that' => 2 );
    PARSE: foreach my $param ( keys %params ) {
        exists $known_options{$param} && 
            do {    #something; 
                    next PARSE; };
        die "Usage: I only know how to 'this' and 'that'\n";
    }
    # Rest of sub here...
}
[download]

Just One Way To Do It.

The values assigned to keys in %known_options are just numeric for this example, but could be coderefs, subrefs, flag values, or whatever you want.

Update: Better late than never, fixed %params = %{ +shift }; syntax.

Dave

Comment on Re: Passing by Named Parameter Select or Download Code

Replies are listed 'Best First'.
I've never seen =that= before! by John M. Dlugosz (Monsignor) on Jan 27, 2004 at 15:41 UTC
`%params = %{ shift };` [download] When I first read it, I thought it implied that shift in list context returns them all and empties the list; that is, gives the same value as `%params= @_;` but also empties out `@_`. After reading Abagail's explaination, I see that he's passing a hashref as the first parameter, rather than passing names/values as individual parameters.	[reply] [d/l] [select]
Re: I've never seen =that= before! by Abigail-II (Bishop) on Jan 27, 2004 at 15:47 UTC
And you don't want to. It means the same as `%params = %shift;` [download] What you might want to see instead is: `%params = %{+shift};` [download] Abigail	[reply] [d/l] [select]