Re: array assignment with list operator , no parenthesis

[Random_Walk]

in reply to array assignment with list operator , no parenthesis

You are not copying an array into your array. You are putting in the result of the operation 7,2,3 which is 7 @one = 7,2,3 which due to precedence is evaluated as (my @one = 7),2,3 as previously explained in this thread. To get the array I think you want in there:

use strict;
use warnings;
use strict;
my @one = (7,2,3);

print "@one\n";

Output :-
7 2 3 

Cheers,
R.

update

shmem pointed out my sloppy summary of this problem. I was concentrating on what I thought he wanted to achieve and lost sight of the true explanation.

[shmem]

You are putting the result of the operation 7,2,3 which is 7.

No, it is not. Consider:

perl -le 'sub foo { 7,2,3 } $foo = foo(); print $foo'
3

It's a precedence problem as other posters pointed out.

update Well, actually... the above example is misleading, since the comma operator is used in list context here - subroutines always return lists (if not declared with a single ($) as prototype, that is). And evaluating a list in scalar context gives its last element - which is just what the comma operator does ;-) A better example might be

print not 1,0;
print $/;
__END__
1

The not operator gets the 0 and turns it to 1.