Re: qw(l i s t) vs (l, i, s, t)
[hildrum:~] tor% perl -v This is perl, version 5.005_03 built for i386-freebsd Thanks guys, newer version is on it's way. Tor
qw(l i s t) vs (l, i, s, t)
Not sure if this belongs here, but anyway:
[hildrum:~] tor% perl -e '@ARGV = qw(1 2); if(scalar(@ARGV)==qw(2 3) )
{print yes\n}'
yes
[hildrum:~] tor%
[hildrum:~] tor% perl -e '@ARGV = qw(1 2); if(scalar(@ARGV)==(2, 3) )
{print yes\n}'
[hildrum:~] tor%
[hildrum:~] tor% perl -e '@ARGV = qw(1 2 3); if(scalar(@ARGV)==(2, 3) )
{print yes\n}'
yes
[hildrum:~] tor%
[hildrum:~] tor% perl -e '@ARGV = qw(1 2 3); if(scalar(@ARGV)==qw(2 3) )
{print yes\n}'
[hildrum:~] tor%
Could someone please explain what is happening here?
I came across it, and had a hard time trying to figure out what was
going on.
Tor
Re: qw(l i s t) vs (l, i, s, t)
hi Could someone please explain what is happening here? I came across it, and had a hard time trying to figure out what was going on. in 5.6.1 it seems to work the correct way, i.e. the first two examples aren't true, while the third and fourth evaluate as true. maybe something was changed in the behaviour of qw, it seems that it in scalar context, it behaves like an array, not like a constant list with the comma operator ... anyway, it sounds more like a bug report thant fwp sc -- stefan stiasny [EMAIL PROTECTED] In an evolving universe, who stands still moves backwards.
Re: qw(l i s t) vs (l, i, s, t)
Tor Hildrum wrote:
Not sure if this belongs here, but anyway:
[hildrum:~] tor% perl -e '@ARGV = qw(1 2); if(scalar(@ARGV)==qw(2 3) )
{print yes\n}'
yes
[hildrum:~] tor%
[hildrum:~] tor% perl -e '@ARGV = qw(1 2); if(scalar(@ARGV)==(2, 3) )
{print yes\n}'
[hildrum:~] tor%
[hildrum:~] tor% perl -e '@ARGV = qw(1 2 3); if(scalar(@ARGV)==(2, 3) )
{print yes\n}'
yes
[hildrum:~] tor%
[hildrum:~] tor% perl -e '@ARGV = qw(1 2 3); if(scalar(@ARGV)==qw(2 3) )
{print yes\n}'
[hildrum:~] tor%
What version of perl are you using? On 5.6.0 and 5.8.0 the first two are false
for me (as I'd expect).
Although thinking about it, qw doesn't have a comma anywhere, so why would it
use the comma operator to return the last element? That wouldn't make much
sense. Would it?
I have no idea. I don't know why I'm bothering to add my 2 pennies worth.
Jasper
--
Lloyd Christmas: Excuse me, miss, what's the soup du jour?
Waitress:The Soup of the Day.
Lloyd Christmas: Mmmm...sounds good, think I'll have that.
Re: qw(l i s t) vs (l, i, s, t)
On Tue, 22 Jul 2003 03:48:53 +0200, [EMAIL PROTECTED] wrote: Could someone please explain what is happening here? I came across it, and had a hard time trying to figure out what was going on. When you use an array in scalar context (e.g. scalar(@ARGV)==foo, or even just @ARGV==foo), it gives the number of elements in the array (either 2 or 3 in your examples). When you use a comma-operator in scalar context (e.g. (2,3)) it throws away the left operand and returns the right one. qw() should act just like a comma'd list, so scalar(qw(2 3)) should be 3. You aren't seeing this, so I suspect you to be using an extremely old version of perl where qw() didn't work as well. If you upgrade to 5.005_55 or higher you should see qw() acting better.
Re: qw(l i s t) vs (l, i, s, t)
Le Mardi 22 Juillet 2003 01:48, Tor Hildrum a écrit : Not sure if this belongs here, but anyway: [snip...] as said Randal : In modern versions of Perl [...] returning either - a list in a list context, - or the last element of a list in a scalar context. (don't forget 'scalar @array' return the lenght of the array) we are in scalar context, because scalar(@ARGV): @ARGV = qw(1 2); # scalar(@ARGV) is 2 (number of elements) scalar qw(2, 3) # is 3, last element of the list = false @ARGV = qw(1 2); # is 2 scalar(2, 3) # is 3 = false @ARGV = qw(1 2 3); # is 3 scalar (2, 3) # is 3 = true @ARGV = qw(1 2 3); # 3 scalar qw(2 3) # is 3 = true This is the result in perl 5.8 on linux so the problem seems in the version of perl you are running, if I understand the answers from Yitzchak Scott-Thoennes, Jasper McCrea and Stefan Stiasny, and obviously Randal. Thanks for my fwp since I have learned more about perl... My apologies for my Frenchglish Jean-Pierre -- Jean-Pierre Vidal
Re: qw(l i s t) vs (l, i, s, t)
John == John Douglas Porter [EMAIL PROTECTED] writes: John Randal L. Schwartz [EMAIL PROTECTED] wrote: ... the last element of a list in a scalar context. John Randal said ...a list in a scalar context. /me faints. I said last element of a list in a scalar context. Don't misparse. There's still no such thing as a list in a scalar context. You can't quote me out of context. And you can't get that context out of me. :-) -- Randal L. Schwartz - Stonehenge Consulting Services, Inc. - +1 503 777 0095 [EMAIL PROTECTED] URL:http://www.stonehenge.com/merlyn/ Perl/Unix/security consulting, Technical writing, Comedy, etc. etc. See PerlTraining.Stonehenge.com for onsite and open-enrollment Perl training!
