One of the points that came up during my talk last night was this bit of code from the golfed version of my Pachinko JAPH:
$#s-(@f=($")x(@s='Just another Perl hacker'=~/./g))%2 @s is being set on the right-hand side of the minus operator, but it's also being used on the left-hand side. Yet the code works. Wha? Here's a simple example that shows the same behavior: DB<1> x $#s - (@s=qw/ 1 2 3 /) 0 '-1' $#s is one less than scalar(@s), so -1 is the correct result. So perl is evaluating the right-hand side before the left-hand side, right? DB<2> x (@t=qw/ 1 2 3 /) - $#t 0 1 Switch the terms around, and it still works! We couldn't figure out the explanation during the meeting... Here's another example, that hints at what is going on: DB<3> x ($#u+0) - (@u=qw/ 1 2 3 /) 0 '-4' Now $#u is being evaluated as 0 rather than 2. So here's what's happening. When perl evaluates $#s - (@s=qw/ 1 2 3 /), it does the following: 1. evaluate $#s 2. evaluate @s=qw/ 1 2 3 / 3. subtract But! when it evaluates $#s, instead of saving the value of $#s, it saves a pointer to the value of $#s. It doesn't follow the pointer until step 3, by which time $#s has been set to 2. Similarly, C<print ++$x, ++$x, ++$x> prints 333 rather than 123, because perl is saving a pointer to the value of $x rather than the actual value of $x. Ronald _______________________________________________ Boston-pm mailing list [email protected] http://mail.pm.org/mailman/listinfo/boston-pm
