Hi Tom!
On Don, 19 Okt 2006, Tom Phoenix wrote:
> >$b = (++$a) + ($a++);
>
> I'm not sure whether C does so, but I believe that Perl does NOT
No, also C does not guarantee any predescribed order. It depends on the
compiler.
BUT: Even if we consider *both* ways of evaluation the expression tree,
we arrive at 8:
Lets take the following evaluation: Let
e1 (++$a)
e2 ($a++)
e3 e1 + e2
we evaluate first e1 and obtain for e1 the value 4 (as it is an
pre-increment), a is now also 4.
Then we evaluate e2: ($a++) which evaluates to 4, too, because this time
it is an post-increment.
Evaluation e3: 4+4 gives 8.
If we do the other ordering:
e1 ($a++)
e2 (++$a)
e3 e1 + e2
we evaluate first expression 1: $a++, which gives 3 (and a is 4 as it is
a post-increment). Then we evaluate e2: which gives 5 (and a is also 5),
so wie have 3 + 5 which is also 8.
I only see an explanation when the pre-increment *NOT*ALWAYS* returns
the value *pre increment.
OR: the expression tree is not actually evaluated in some sub-tree
order, but in some leave up order. Then we would have
e1 ++$a
e2 $a++
e3 (++$a)
e4 ($a++)
e5 e3 + e4
which would yield:
e1 4 pre increment, a=4
e2 4 a=5
e3 4
e4 5 it returns the current value of a
e5 9
(the above interpretation is very unclear, but it is the only way I can
see an explanation)
But this would be extremely counter-intuitive!!!
> Did this come up in a real-world situation, or were you specifically
> seeking to test the limits of Perl? In other words, what problem, if
> any, are you trying to solve?
Teaching Perl to students, and to get them to understand the difference
between pre and post increment.
Best wishes
Norbert
-------------------------------------------------------------------------------
Dr. Norbert Preining <[EMAIL PROTECTED]> Università di Siena
Debian Developer <[EMAIL PROTECTED]> Debian TeX Group
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-------------------------------------------------------------------------------
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