The "fix" function converts a real number to the nearest smaller number.
How to round a real number to the nearest larger number?
Thanks in advance.
; Rounding function  Doug Broad  02/19/2003
; Additional credits: Joe Burke, Peter Toby
; Original name: (defun Round (value to) ...)
;
; Examples:
; (ALE_Number_Round 1.11111 1) => 1
; (ALE_Number_Round 1.11111 0.1) => 1.1
; (ALE_Number_Round 1.11111 0.01) => 1.11
; (ALE_Number_Round 1.11111 0.001) => 1.111
; (ALE_Number_Round 1.11111 0.0001) => 1.1111
; (ALE_Number_Round 1.11111 0.0000001) => 1.11111
;
; (ALE_Number_Round 1.565656 10) => 0
; (ALE_Number_Round 1.565656 1) => 2
; (ALE_Number_Round 1.565656 0.1) => 1.6
; (ALE_Number_Round 1.565656 0.01) => 1.57
; (ALE_Number_Round 1.565656 0.001) => 1.566
; (ALE_Number_Round 1.565656 0.0001) => 1.5657
; (ALE_Number_Round 1.565656 0.00001) => 1.56566
; (ALE_Number_Round 1.565656 0.0000001) => 1.56566
;
(defun ALE_Number_Round (ImpVal To_Val)
(setq To_Val (abs To_Val))
(*
To_Val
(fix
(/
(
(if (minusp ImpVal)

+
)
ImpVal
(* To_Val 0.5)
)
To_Val
)
)
)
)
@Marc'Antonio AlessiThanks for pointing that. :'(
To be exact: :police:
Your function rounds to infinity and not to the nearest largest number.
(ALE_NUMBER_ROUND 1.5 1) => 2
$ (type (round 2.49))
INT
$ (defun round (val) (float (fix (+ val 0.49999999999999994))))
ROUND
_$ (type (round 2.49))
REAL
_$ (round 1234567890)
1234567890
_$ (round 12345678901)
1.23457e+010
(defun round (num pre / b)
;; Enhanced rounding
;; num  a number
;; pre  precision , type in an interg like 1 0 2 ...
;; (round 1.23456789 3) > 1.235
;; (round 1.23456789 0) > 1.0
;; (round 123456789 3) > 1.23457e+008
;; (round 98764567 5) > 9.88e+007
(setq b (expt 10.0 pre))
(/ (fix (+ (* num b) (* 0.5 (sign num)))) b)
)
(defun sign (x) (cond ((minusp x)1.)(1.0)))
(round 1.9) => 1
What is your rounding rule?