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
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