Without the subtraction, all perl needs to do is remember where the value of $ofs is, and get it when it's time to output the resulting string. Of course, in between, the value gets modified. With the subtraction, perl gets the value of $ofs, subtract 0, and scribbles away the result. That result won't get modified when $ofs gets modified.

Yep. Thats what I tried to say myself, but youv'e done a much better job. :-)

Don't modify a variable and use its value elsewhere in the same expression. Don't modify a variable twice in the same expression. Don't assume Perl has a defined order of evaluation.

Yep, its just that Perls loose rules for this stuff make it so easy to do. Given enough rope to hang themselves... :-)

print 5+1;
[download]

update: I see that I was confusing precedence and evaluation order.

Perl does have a defined order of evaluation;

Do you know something that neither p5p nor comp.lang.perl.misc knows? Order of evaluation has been discussed recently on both forums, and noone one these forums has been able to show that Perl has a defined order of evaluation. Please quote the relevant parts of the documentation that define order of evaluation in general. (Order of evaluation is defined for &&, ||, and, or and comma in scalar context, but not generally).

How can I trust

print 5+1;
[download]

always prints 6, and not 5 (and a warning)?

Precedence garantees that. Just like C, Perl does have well defined precedence and associativity rules. But neither language has a defined order of evaluation.

Abigail

This one thing amazes me. ML is functional (though not purely), side-effects are pretty rare there, yet it does specify the order of evaluation in all cases. C and Perl are full of side-effects, yet they don't.

Jenda
Always code as if the guy who ends up maintaining your code will be a violent psychopath who knows where you live.
   -- Rick Osborne

Edit by castaway: Closed small tag in signature