Re: what happens with $b = $b++?

[Michael Turner]

>     Michael> $b = 1; $b = $b++;
>     Michael> Result:
>     Michael> $b = 1
>
> $b++ is synonymous with $b = $b + 1.  So, your second statement reads:
>
> '$b = $b = $b + 1'.
>
> If you replace your code with:
>
> '$b = 1; $b++;' or '$b = 1; $b = $b + 1', you'll get more expected
> behaviour.  :)
>
> - Chris.
> --
>


Thanks for the response, Chris,

but check this out:

$b=1;
$b = ++$b;
result: 2

So, that isn't the whole story. This has to do with pre- & post 
"timing", I think.

/Michael Turner


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