'Twas brillig, and Andy McKenzie at 24/08/10 21:42 did gyre and gimble: > Even if I'd thought about it in terms of the architecture, I > would have assumed that PHP would treat a two-bit number as a two-bit > number You two-bit hustler! In all seriousness tho', where do you ever provide a two bit number? $n = 2; The above is a number, it's not implicitly a two bit number... I mean, if I do: $n = 2; $n += 2; What do I get then? $n == 0? That's what *should* happen if $n was indeed two bits. It would overflow and wrap around. Your number in $n is in fact represented by whatever the variable type is. In this case it's a 32 bit integer number. Col -- Colin Guthrie gmane(at)colin.guthr.ie http://colin.guthr.ie/ Day Job: Tribalogic Limited [http://www.tribalogic.net/] Open Source: Mandriva Linux Contributor [http://www.mandriva.com/] PulseAudio Hacker [http://www.pulseaudio.org/] Trac Hacker [http://trac.edgewall.org/] -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
