in reply to my sub ?

You could make anonymous subrouting references to (x) in sub a & b like so: 
sub a {
        my $ref = sub {
                print "x1\n";
        };

        &$ref;
}

sub b {
        my $ref = sub {
                print "x2\n";
        };

        &$ref;
}

&a;
&b;

[download]
The only thing that will look different is the call to the fucntion if you can live with that.

/\/\averick
perl -l -e "eval pack('h*','072796e6470272f2c5f2c5166756279636b672');"

Replies are listed 'Best First'.
Re: Re: my sub ?
by shotgunefx (Parson) on Mar 25, 2002 at 20:00 UTC
    Doesn't calling a sub reference with & imply the passing of your @_ ?

    -Lee

    "To be civilized is to deny one's nature."
[reply]
      Yes it does, but if you just stick $ref; out there by itself, or with () after it, perl doesn't get that that's supposed to be a function call and throws a syntax error.

      You pretty much have to have the & as far as I know.

      Or you could use $ref->(); It's monday and I'm being a blonde.

      Update:Thanks chromatic, Fletch, and shotgunefx for the clue-by-four-ing.

      /\/\averick
      perl -l -e "eval pack('h*','072796e6470272f2c5f2c5166756279636b672');"

[reply]

        $ref->() works just fine. As would &{$ref}().

[reply]
[d/l]
[select]
        Unless I want the behavior, I usually call a code ref with
        $ref->();
        Probably 9 out of 10 times you can get away with it but it can product subtle bugs. I wasn't originally aware of the implied passing of @_ and it was never a problem.. until.

        -Lee

        "To be civilized is to deny one's nature."
[reply]
[d/l]

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