Re: swapping vars, two ways, 2 results...

[pg]

By doing

my ($b, $a) = ($a, $b)

you are not swaping, you simply defined another set of variables. Try this, it will clearly show you what happend, and why it is way fast:

$a = 100;
$b = 200;
my ($a, $b) = ($b, $a);
print $a, "\n"; #200
print $b, "\n"; #100
print $main::a, "\n";#100
print $main::b, "\n";#200

Update:

When you do:

($b, $a) = ($a, $b)

You are not working against two scalars, in stead you are working against two arrays, and two scalars. Not that it is slow, I don't even think it take less memory as you thought. Although you are not using any temps, Perl is using.

[Sihal]

well it does use less memory, but that may be due to the length of the vars...
Anyway, this was a kinda stupid post but I wanted to write it cause I saw in one of my team mates code that he would never swap perls way, but rather Cs way, using a temp var. I first thought why wrinting it Cs way when you can use a more compact form. But then I saw that in some case it was actually totally different both in terms of cpu usage and mem, hence the post.

[Sihal]

but why is it that much faster ?

[BrowserUk]

This ($a,$b) = ($b,$a); is notionally equivalent to

{
    local @list;
    $list[0]=$b;
    $list[1]=$a;
    $a=$list[0];
    $b=$list[1];
    undef @list;
}

Instead of allocating one temprary scalar and doing 3 assignments, it is allocating a temporary list, performing 4 assignments and deallocating the list.

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