On Oct 2, 2009, at 15:22, Daevid Vincent wrote:
-----Original Message-----
From: Ben Dunlap [mailto:bdunlap@xxxxxxxxxxxxxxxxxx]
Sent: Friday, October 02, 2009 2:58 PM
To: php-general@xxxxxxxxxxxxx; Daevid Vincent
Subject: Re: Whacky increment/assignment logic with
$foo++ vs ++$foo
mind-blowing. What the heck /is/ supposed to happen when
you do this:
$a = 2;
$a = $a++;
echo $a;
Seems like any way you slice it the output should be 3. I
guess what's
... and, in fact, that /is/ how C behaves. The following code:
int a = 2;
a = a++;
printf("a = [%d]\n", a);
Will output "a = [3]". At least on Ubuntu 9 using gcc 4.3.3.
So I retract my initial terse reply and apologize for
misunderstanding
your question.
Ben
EXACTLY! :)
God (or diety of your choice) bless you for "getting" what I'm
saying and
proving that it's not "C" like either. That just adds credence to
my/our
argument.
d
I think if you rewrote the above example as
int a = 2;
b = a++;
printf("b = [%d]\n", b);
"b" would be 2 when printed. However, after the second line (b = a+
+;) finished executing, "a" would then be 3.
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