08-19-2003, 03:04 AM
Take a closer peek at neo's example:
this is the equivalent of Log 2 n, the solution for 2^x = n. if you say: int(log(number&) / log(2)) you get the number of digits.
so if you're using a dynamic array:
dim scraparray(200) as integer '<- oughta be enough
'do your thing
redim myarray(int(log(number&) / log(2))
then copy your array.
of course, an easier way to do it would be:
and then curdigit is the number of digits used.
and to whitetiger: you could do that conversion 1000 times in your program, and some of those things I did roll over into other areas of programming, so it's a good habit to use them.
EDIT: nuts, I misunderstood your question. BRB
Code:
LOG(number&) / LOG(2)
this is the equivalent of Log 2 n, the solution for 2^x = n. if you say: int(log(number&) / log(2)) you get the number of digits.
so if you're using a dynamic array:
dim scraparray(200) as integer '<- oughta be enough
'do your thing
redim myarray(int(log(number&) / log(2))
then copy your array.
of course, an easier way to do it would be:
Code:
cudigit = 0
while number& <> 0
if number& and 1 then array(curdigit) = 1 else array(curdigit = 0
number& = number& \ 2
curdigit = curdigit + 1
wend
and then curdigit is the number of digits used.
and to whitetiger: you could do that conversion 1000 times in your program, and some of those things I did roll over into other areas of programming, so it's a good habit to use them.
EDIT: nuts, I misunderstood your question. BRB
i]"I know what you're thinking. Did he fire six shots or only five? Well, to tell you the truth, in all this excitement, I've kinda lost track myself. But being as this is a .44 Magnum ... you've got to ask yourself one question: 'Do I feel lucky?' Well, do ya punk?"[/i] - Dirty Harry