On Tue, Sep 22, 2009 at 11:13, Telemachus <[email protected]> wrote:
> On Tue Sep 22 2009 @ 10:56, Chas. Owens wrote:
>> On Tue, Sep 22, 2009 at 09:40, Bryan R Harris
>> <[email protected]> wrote:
>> > Can you explain how perl interprets this? I would've incorrectly thought:
>> >
>> > 1) "() = func_that_return_list_value" tries to assign the list to "()"
>> > which perl would complain about.
>> >
>> > 2) Then assign the result to $size, which it wouldn't like either.
>> >
>> > Just when you think you're starting to understand perl...
>> snip
>>
>> The number of "catcher" elements in a list does not need to be equal
>> to the number of "thrown" items:
>>
>> my ($x, $y) = (1, 2, 3);
>>
>> List assignment behaves differently in list and scalar contexts. In
>> list context, it returns the list that was successfully assigned to
>> variables. In scalar context, it returns the number of items that
>> tried to be assigned. So, in
>>
>> perl -le '$z = ($x, $y) = (qw/a b c/); print "$x, $y, $z"'
>>
>> $z is 3 (because there were three items in RHS of the assignment), $x
>> is "a", and $y is "b". "c" is silently discarded.
>
> All of this makes sense, but I have to admit I find it odd that the
> following doesn't even produce a "void context" warning:
>
> my @array = (1, 2, 3);
> my $num = () = @array;
> print $num, "\n";
snip
Ah, but it isn't a "Useless use of (%s) in void context". Even if "my
$num =" weren't there it wouldn't be useless. Examine this code:
#!/usr/bin/perl
use strict;
use warnings;
sub foo {
my $want = wantarray;
print defined $want ? $want ? "list" : "scalar" : "void", "\n";
}
foo;
scalar foo;
() = foo;
--
Chas. Owens
wonkden.net
The most important skill a programmer can have is the ability to read.
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