in reply to How does this code work? (from "Perl is Unix")

run_fork and child are function calls declared with the & prototype (See perlsub: Prototypes). So, yes, they auto-subify their first argument (as though you had called run_fork(sub { ... })).

Good Day,
    Dean

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Re^2: How does this code work? (from "Perl is Unix")
by Anonymous Monk on Nov 06, 2009 at 15:18 UTC

    Ohhhhhh. I see. Thank you for pointing that out! I've avoided using Prototypes in my own code, and had almost forgotten about them.

    FWIW, I much prefer the more explicit syntax that you gave as an equivalent example. That is, something like this seems sensible and obvious to me:

    #!/usr/bin/env perl

use strict;
use warnings;

sub call_it {
    my $fn_ref = shift;
    &{$fn_ref}();
}

call_it( sub { print "hi\n"  } );
call_it( sub { print "bye\n" } );

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[d/l]

      If you understand map { foo($_) } then this code shouldn’t give you any trouble.

      Makeshifts last the longest.

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[d/l]

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