On Fri, Dec 06, 2002 at 03:30:35PM -0800, david wrote:
> Paul Johnson wrote:
>
> > But it might. The behaviour is undefined. The compiler may do as it
> > will. Google for "sequence point" if you want to find out more.
> >
> > The behaviour in Perl is undefined too, but more in the sense that
HAHAHAHAHAHA!
-Chris
On Friday 06 December 2002 03:51 pm, [EMAIL PROTECTED] wrote:
> He forgot to mention the:
>
> use Advil;
>
> $pills = new Advil(2);
> unless($pills->take(orally)) {
> sleep 40;
> }
>
> http://danconia.org
>
>
>
Paul Johnson wrote:
> But it might. The behaviour is undefined. The compiler may do as it
> will. Google for "sequence point" if you want to find out more.
>
> The behaviour in Perl is undefined too, but more in the sense that the
> behaviour has not been defined rather than that the behaviour
On Fri, Dec 06, 2002 at 02:46:13PM -0800, david wrote:
> btw, the ++$i / ++$i gives you a 1 thing behaves differently in other
> programming languages. For example, try the following in C++:
>
> #include
> void main{
> int i=2;
> int j=++i/++i;
> cout< }
>
> won't give y
Jeff 'Japhy' Pinyan wrote:
>
> Strictly speaking, there is another major difference no one has mentioned
> yet (and that many people might have trouble understanding). Using
> $count++ returns a NUMBER OR STRING, and then increments $count's value.
> ++$count increments $count's value, and retur
On Fri, Dec 06, 2002 at 04:22:19PM -0500, Jeff 'japhy' Pinyan wrote:
> On Dec 6, Paul Johnson said:
>
> >On Fri, Dec 06, 2002 at 11:58:37AM -0500, Danny Miller wrote:
> >
> >> Strictly speaking, ++$count is faster than $count++.
> >
> >Strictly speaking, perl will convert $count++ to ++$count if i
He forgot to mention the:
use Advil;
$pills = new Advil(2);
unless($pills->take(orally)) {
sleep 40;
}
http://danconia.org
On Fri, 6 Dec 2002 16:22:19 -0500 (EST), "Jeff 'japhy' Pinyan" <[EMAIL PROTECTED]>
wrote:
> On Dec 6, Paul Johnson sai
On Dec 6, Paul Johnson said:
>On Fri, Dec 06, 2002 at 11:58:37AM -0500, Danny Miller wrote:
>
>> Strictly speaking, ++$count is faster than $count++.
>
>Strictly speaking, perl will convert $count++ to ++$count if it can.
Strictly speaking, there is another major difference no one has mentioned
y
On Fri, Dec 06, 2002 at 11:58:37AM -0500, Danny Miller wrote:
> Strictly speaking, ++$count is faster than $count++.
Strictly speaking, perl will convert $count++ to ++$count if it can.
--
Paul Johnson - [EMAIL PROTECTED]
http://www.pjcj.net
--
To unsubscribe, e-mail: [EMAIL PROTECTED]
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Well, it depends how you use them.
Strictly speaking, ++$count is faster than $count++.
say $count = 5
$num1 = $count++; #$num1 would = 5 and $count would = 6
$num2 = ++$count; #$num2 and $count would equal 6
Regards,
Danny
-Original Message-
From: Mystik Gotan [mailto:[EMAIL PROTECTE
Yup... good catch... i should run my code rather than just type it in a
mail window i guess :-)
> Don't you mean ++$count??
>
> > my $i = $++count;
> > print "$i\n";
> >
> > On Sat, 2002-12-07 at 03:54, Mystik Gotan wrote:
> > > Hiya,
> > >
> > > is there any difference between $count++
Gah!
bassackwards again :(
-Original Message-
From: LRMK [mailto:[EMAIL PROTECTED]]
Sent: Friday, December 06, 2002 12:11
To: Yacketta, Ronald
Cc: [EMAIL PROTECTED]
Subject: Re: Difference between $count++ and ++$count
++$count will increment $count before it is used
$count++ will
: Friday, December 06, 2002 11:02 PM
Subject: RE: Difference between $count++ and ++$count
> Yes
>
> If I recall correctly:
>
> ++$count will increment $count after it is used
> $count++ will increment $count before it is used
>
> -Ron
>
> -Original Message-
> F
Yes sir there is! And it's quite a goober if you ask me, but very useful.
I'll show the difference by example.
my $SOME_CONSTANT = 2; # NO MAGIC NUMBERS! (hehe)
my $pre_increment = 0;
my $post_increment = 0;
my $pre_result = ++$pre_increment + $SOME_CONSTANT;
my $post_result = $post_i
On Friday 06 December 2002 11:09 am, simran wrote:
> Yup...
>
> In program 1 you will get $i to be 1 and then $count will be set to 2
> In program 2 you will get $count to be set to 2 and then assigned to $i
> so now $i will also be 2.
>
> Its just a prcedence thing...
>
> Program 1
> --
$count++ - will display value then add
++$count - will add then display
as in where $count = 5;
printf "%5d\n", $count++ ;
would display 5 and $count would be 6
printf "%5d\n", ++$count ;
would add then display 6
Yup...
In program 1 you will get $i to be 1 and then $count will be set to 2
In program 2 you will get $count to be set to 2 and then assigned to $i
so now $i will also be 2.
Its just a prcedence thing...
Program 1
--
my $count = 1;
my $i = $count++;
print "$i\n";
Program 2
Yes it is a differense.
code:
#!/usr/bin/perl -w
use strict;
my $hepp = 3;
my $hopp;
$hopp =++$hepp;
print "hopp: $hopp\n";
$hepp = 3;
$hopp = $hepp++;
print "hopp: $hopp\n";
Mystik Gotan wrote:
>
> Hiya,
>
> is there any difference between $count++ and ++$count?
> Just wondering.
>
>
Yes $count++ is incremented afterwards where ++$count is incremented
first
$a=1
C$=++a$ #c$ gets 2
C$=a$++ #c$ gets 1 and then a$ is increment to 2
Print "$c $a" #returns "1 2".
Paul
> -Original Message-
> From: Mystik Gotan [mailto:[EMAIL PROTECTED]]
> Sent: Friday, December 06, 2002 11
Yes
If I recall correctly:
++$count will increment $count after it is used
$count++ will increment $count before it is used
-Ron
-Original Message-
From: Mystik Gotan [mailto:[EMAIL PROTECTED]]
Sent: Friday, December 06, 2002 11:54
To: [EMAIL PROTECTED]
Subject: Difference between $coun
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