Hi,

>>Just tested (you never know with computers): bitand(2^32, 2^32) returns 0 since we are exceeding 32 bits (2^31 ..... 2^0)
<<

Yea, you're right - realised my example was only 24 bit straight after posting. So how about, for x and y:

nresult = BITAND(int(x/0xffffffff),int(y/0xffffffff))* 0xffffffff ;
+ BITAND(Mod(x,0xffffffff),mod(y,0xffffffff))


Regards,
Viv