re:"It seems that '$b = $b++;' is always a coding mistake. Nobody
should use that syntax on purpose"
True, True. I can't see why you would ever choose to use that to get
$b to increment itself...just use "$b++" or "++$b" depending on the
effect you want!
I appreciate everyone who wrote, but I remain confused.
It seems to me that there is a weird short-circuit in the following code:
$b = 1;
$b = $b++;
print $b;
--> 1
I expected "2"
Whereas,
$b = 1
$b = ++$b;
print $b;
--> 2
As I expected.
I translate the autoincrement ($b++) as $b + 1, but post-increment
doesn't seem to work when assigned to itself, whereas pre-increment
does. So I was hoping for some confirmation that it is supposed to
work that way, and perhaps why...
So while Connie & Dave & Chris have answered the question to a
degree (and I thank them each), they haven't resolved my confusion.
It seems that '$b = $b++;' is always a coding mistake. Nobody should
use that syntax on purpose, apparently, because it ignores the 'side
effect' auto-increment. Is this true? I'll try to write my questions
better in the future.
/Michael Turner
--
Dave Tenen
basicSolutions?
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Coming to you from Spec Pond, and I swear that fish
was [_______________________________________] that big!!!
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