Re: simple eurphoria question
- Posted by Chris Bensler <bensler at nt.net> Dec 09, 2006
- 696 views
Mark K. Akita wrote: > > Chris Bensler wrote: > > > > Mark K. Akita wrote: > > > > > > Hi John, > > > > > > This function seems to work as well for converting atom to integer. > > > > > > }}} <eucode> > > > > > > function int(atom a) > > > > > > return floor(a)+(a<0) > > > > > > end function > > > > > > </eucode> {{{ > > > > > > Good luck! > > > > > > Mark K. Akita > > > <a > > > href="http://marksarts.blogspot.com/">http://marksarts.blogspot.com/</a> > > > > That's what I thought Mark, but it's incorrect. :) > > Try with this set of numbers: 1, 1.1, 1.9, 2, -1, -1.1, -1.9, -2 > > > > Mike Nelson, or Larry Miller's solutions are correct. > > > > Chris Bensler > > ~ The difference between ordinary and extraordinary is that little extra ~ > > <a href="http://empire.iwireweb.com">http://empire.iwireweb.com</a> - Empire > for Euphoria</font></i> > Good catch Chris, > OK.. how about this function instead > > }}} <eucode> > function int(atom a) > > return a-remainder(a,1) > > end function > </eucode> {{{ > > This should work with negative numbers that already are integers. > > Mark K. Akita > <a href="http://marksarts.blogspot.com">http://marksarts.blogspot.com</a> It's interesting that works. I would have expected the result of anything divided by 1 to be equal to itself (1.2/1 = 1.2). Apparently remainder() does integer division (1.2/1 = 1). Chris Bensler ~ The difference between ordinary and extraordinary is that little extra ~ http://empire.iwireweb.com - Empire for Euphoria