At 12:08 PM 6/30/02 -0400, Michael Turner wrote:
>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.
It depends on the order things take place in Perl.
$b = $b++
means, take the current value of $b (1), assign it to $b, and when
you're done with the expression, increment the value of $b by 1. There
are two assignments to $b going on there: the explicit one and the
implicit one. Since they both take place after everything else has
happened, it ought not to be suprising that they might take place in
any order. Here's what the ops executed look like:
$ perl -Dt -le '$b=$b++'
EXECUTING...
(-e:0) enter
(-e:0) nextstate
(-e:1) gvsv(main::b)
(-e:1) postinc
(-e:1) gvsv(main::b)
(-e:1) sassign
(-e:1) leave
What I think this means is, "Take the current value of $b, save it as
the result of the expression. Increment $b. Set $b to the saved
result of the expression."
Whereas the other case is much easier to explain:
$b = ++$b
means, increment the value of $b, take the resulting value as the
result of the expression, assign it to $b.
Of course, "Don't do that." But since you asked...
--
Peter Scott
Pacific Systems Design Technologies
http://www.perldebugged.com/
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