List assignment in scalar context

Grimnebulin has asked for the wisdom of the Perl Monks concerning the following question:

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•Re: List assignment in scalar context by merlyn (Sage) on Jan 10, 2005 at 18:45 UTC
"split" has some DWIM hackery to ensure that it doesn't work too hard, so it splits to one more than the container on the left, in the absence of a third-parameter hint. -- Randal L. Schwartz, Perl hacker Be sure to read my standard disclaimer if this is a reply.	[reply]
Re^2: List assignment in scalar context by Grimnebulin (Initiate) on Jan 10, 2005 at 19:05 UTC
Aha. Guess I should have RTFM. It contradicts my usually-accurate intuition about how list/scalar context works in Perl, so I hope I can be forgiven. My real-life code that inspired this question goes something like this: `my $count = my ($arg1, $arg2, $arg3) = split /\s+/, $argstring; $count == 3 or die "3 arguments expected; you supplied $count\n";` [download] So if I want the real argument count, I have to split into a temporary array first, or else use a dummy trailing array in the assignment list. Gross. I can't say I care for this feature. Smells like premature optimization to me. And I notice it doesn't work across function calls. That could be a nasty surprise for someone someday.	[reply] [d/l]
Re^3: List assignment in scalar context (disable the DWIM) by tye (Sage) on Jan 10, 2005 at 20:00 UTC
`my $count = my ($arg1, $arg2, $arg3) = split /\s+/, $argstring, -1; # ^^^^` [download] - tye	[reply] [d/l]
Re^3: List assignment in scalar context by demerphq (Chancellor) on Jan 10, 2005 at 19:32 UTC
So if I want the real argument count, I have to split into a temporary array first, or else use a dummy trailing array in the assignment list. Gross. Personally i think the issue here is that what you are trying to do is much less common than the opposite. Ie, more people would rather have split// be faster on spliting a few elements out scenarios than they would with have it be slower but more consistant with other list returning functions. After all, split may be quite the wrong way to do this. Arguably `my $count=()=$argstring=~/\s+/g; my ($arg1, $arg2, $arg3) = split /\s+/, $argstring;` [download] is better anyway, as if the $argstring is long and you only are going to store three elements then its better not do the actual splitting (creating new svs, populating them etc) when you dont need to. --- demerphq	[reply] [d/l]
Re^2: List assignment in scalar context by dragonchild (Archbishop) on Jan 10, 2005 at 18:50 UTC
That doesn't explain why the following happens: `my ($a,$b,$c); my $n1 = ($a,$b,$c) = split /,/, "1,2,3,4,5"; my $n2 = my @x = ($a,$b,$c) = split /,/, "1,2,3,4,5"; print "$n1 ... $n2\n"; ---- 4 ... 3` [download] If it splits to one more, then why is that one going across one boundary, but not two? Being right, does not endow the right to be rude; politeness costs nothing. Being unknowing, is not the same as being stupid. Expressing a contrary opinion, whether to the individual or the group, is more often a sign of deeper thought than of cantankerous belligerence. Do not mistake your goals as the only goals; your opinion as the only opinion; your confidence as correctness. Saying you know better is not the same as explaining you know better.	[reply] [d/l]
Re^3: List assignment in scalar context by !1 (Hermit) on Jan 10, 2005 at 19:21 UTC
Oh come off it. `sub list { return 1..5 } my ($a,$b,$c); my $n1 = ($a,$b,$c) = list(); my $n2 = my @x = ($a,$b,$c) = list(); print "$n1 ... $n2\n"; ---- 5 ... 3` [download] When you have `scalar = list = list`, you will always get the count of items in the rightmost list in scalar. In your $n2 example, your array is getting three items from the list. It isn't a scalar, so it isn't getting the count. What else would be expected? The behavior that merlyn is referring to is the reason why you don't see the complete count in the scalar.	[reply] [d/l] [select]
Re^4: List assignment in scalar context by dragonchild (Archbishop) on Jan 10, 2005 at 19:25 UTC
Re^5: List assignment in scalar context by !1 (Hermit) on Jan 10, 2005 at 19:32 UTC
Re^5: List assignment in scalar context by jonadab (Parson) on Jan 11, 2005 at 00:15 UTC
Re^3: List assignment in scalar context by Aristotle (Chancellor) on Jan 10, 2005 at 22:35 UTC
You tricked yourself. In the first case, the list assignment is evaluated in scalar context. In the second case, the list assignment is evaluated in list context and assigned to `@x`, yielding an array with three elements, which is then evaluated in scalar context. And an array with three elements evaluates to 3 in scalar context. Makeshifts last the longest.	[reply]