Re: 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
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Cheers,
R.

Pereant, qui ante nos nostra dixerunt!

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.

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Re^2: array assignment with list operator , no parenthesis by shmem (Chancellor) on May 14, 2007 at 14:10 UTC
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 $foo' 3` [download] It's a precedence problem as other posters pointed out. update Well, actually... the above example is misleading, since the comma operator is used in list context here - subroutines always return lists (if not declared with a single ($) 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 $/; __END__ 1` [download] The not operator gets the `0` and turns it to `1`. --shmem _($_=" "x(1<<5)."?\n".q·/)Oo. G°\ / /\_¯/(q / ---------------------------- \__(m.====·.(_("always off the crowd"))."· ");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}	[reply] [d/l] [select]

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Re^2: array assignment with list operator , no parenthesis
by shmem (Chancellor) on May 14, 2007 at 14:10 UTC

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
[download]

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
[download]

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

--shmem

_($_=" "x(1<<5)."?\n".q·/)Oo.  G°\        /
                              /\_¯/(q    /
----------------------------  \__(m.====·.(_("always off the crowd"))."·
");sub _{s./.($e="'Itrs `mnsgdq Gdbj O`qkdq")=~y/"-y/#-z/;$e.e && print}

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