On Fri, Dec 06, 2002 at 03:30:35PM -0800, david wrote: > Paul Johnson wrote: > > > But it might. The behaviour is undefined. The compiler may do as it > > will. Google for "sequence point" if you want to find out more. > > > > The behaviour in Perl is undefined too, but more in the sense that the > > behaviour has not been defined rather than that the behaviour has been > > defined as undsefined. > > > > True, maybe. > > But I don't know what you mean by the behaviour in Perl is undefined. > To me, the behaviour is well defined (ie, you won't find an exception where > the above bahaviour will behaves differently). it's simply $i / $i which > always gives you a 1 in Perl except where $i = 0 > david
A couple of years ago, someone not too far away, with a penchant for starting these kind of threads, asked about: $x = 10; print ++$x / $x--; # .909090 (or 10/11) Jarkko Hietaniemi, who was in charge of creating the 5.8 release, which was in development at the time, said: Yes, but *what* output do you expect? I think whichever of the 11/10, 11/9, 10/9, 1, 10/11, 9/11, 9/10, 42, or seven pink elephants dancing flamenco on a pinhead you choose, I can argue otherwise. You say the dividend must be evaluated first, I say the divisor, and so on. I wouldn't rely on the order of evaluation of operands. It might change. At the very least it will be confusing to anyone looking at it. -- Paul Johnson - [EMAIL PROTECTED] http://www.pjcj.net -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
