Re: Parameter passing

Why pass references in - why not just pass the scalars $x and $y?

my ($x, $y);
if (fn($x, $y)) {
    die "fn() failed\n";
}

# do stuff with $x and $y...
exit 0;

sub fn
{
    my ($fn_x, $fn_y) = @_;
    ($fn_x, $fn_y) = ('x', 'y');

    my $rc = 0;
    $rc = 1 if $fn_x ne 'x'; # never happens, just an example

    return 0;
}
[download]

The only change I made was to change the call - instead of passing references in, just pass the scalars. You aren't changing the values of the variables in the subroutine - I can't see any reason why you would want to pass the references instead of the actual scalar values.

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Re: Re: Parameter passing by gjb (Vicar) on Oct 30, 2003 at 16:11 UTC
Try running this: `#!/usr/bin/perl use strict; my $a = 2; f($a); print "$a\n"; sub f { my $x = shift; $x = 5; }` [download] You'll notice it prints '2', while according to you it should print '5'. This is precisely why a reference should be passed to the function. If you don't, you actually pass a value (2) and initialize the local variable `$x` to it. Later in the sub the value of `$x` is change to 5. Now why should that change the value of `$a` in the main program? It shouldn't since the subroutine only knows about the value of `$a` which happens to be 2, not about `$a` itself so the value of `$a` will remain 2 as it was before the function call. Changing the code above to: `#!/usr/bin/perl use strict; my $a = 2; f(\$a); print "$a\n"; sub f { my $x = shift; $$x = 5; }` [download] will actually print '5' rather than '2'. Now you pass the address of the variable `$a` to the sub and it gets stored in `$x`. The value at the address in `$x` is set to 5 by dereferencing `$x` with `$$x`. Since `$x` is the address of `$a`, it's value is indeed changed to 5. Hope this helps, -gjb-	[reply] [d/l] [select]

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Re: Re: Parameter passing
by gjb (Vicar) on Oct 30, 2003 at 16:11 UTC

Try running this:

#!/usr/bin/perl

use strict;

my $a = 2;
f($a);
print "$a\n";

sub f {
    my $x = shift;
    $x = 5;
}
[download]

This is precisely why a reference should be passed to the function. If you don't, you actually pass a value (2) and initialize the local variable $x to it. Later in the sub the value of $x is change to 5. Now why should that change the value of $a in the main program? It shouldn't since the subroutine only knows about the value of $a which happens to be 2, not about $a itself so the value of $a will remain 2 as it was before the function call.

Changing the code above to:

#!/usr/bin/perl

use strict;

my $a = 2;
f(\$a);
print "$a\n";

sub f {
    my $x = shift;
    $$x = 5;
}
[download]

$a

$x

$$x

$x

$a

Hope this helps, -gjb-

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