Re: Evaluation of ++ different in C and perl?

[Norbert Preining]

Dr.Ruud wrote:

$ perl -wle '
  $a = 3;
  $b = 0 + (++$a) + ($a++);
  print "b=$b\n";
'
b=8
:)

Nup, this is not the solution:

$a = 3;
$b = 0;
$b = 0 + (++$a) + ($a++);
$c = 3;
$d = 0;
$d =     (++$c) + ($c++);
print "b=$b\n";
print "d=$d\n";

prints:
b=8
d=9

Don't tell me that in perl
  0 + somethingnumerical != somethingnumerical

So it is NOT dependent on the inizialization!

Ciao

Norbert

--
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
<http://learn.perl.org/> <http://learn.perl.org/first-response>