General information
Category:
Coding, syntax & commands
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
Previous
Next
Reply
View the map of this thread
View the map of this thread starting from this message only
View all messages of this thread
View all messages of this thread starting from this message only