How to round float numbers to get this result?
0.45 == 0
0.85 == 0
or
4.25 == 0
6.25 == 10
or
45.5 == 50
44.5 == 40
–this will do it , but is to slow
fn roundLastInt val =
(
fn roundInteger intg = if intg > 4 then 10 else 0
val = val as integer -- cut float values if exist
str = val as string
if str.count == 1 then -- if value si between 0 to 9
(
return roundInteger val
)
else if str.count > 1 do -- if value is between 10 to infinite
(
new_str = substring str 1 (str.count-1) --cut the last int
last_int = roundInteger (str[str.count] as integer) --round last int
return ((new_str+"0") as integer + last_int) as integer -- join last int back
)
)
You can use “Ceil” and “Floor” to get the nearest whole numbers to a given float.
Hi Marco Brunetta
how say ceil to get value 45.257 and return 50 and not 46.0 ?
and how to say floor to get a value 41.257 and return 40 not 41.0 ?
i tried a lot of methods and here is a fastest and cheapest:
fn roundFloat d pre:0.1 =
(
d = (d as float)/pre
v = if (d - (v1 = floor d)) > ((v2 = ceil d) - d) then v2 else v1
v*pre
)
Hi Denis :wavey:
with you function I cant get right result
roundFloat 45.258 pre:1
return 45.0 not 50 :-)))
Is not a typical round…
What I want is to round it much more or much less :-))))
–something like this should be done for last value in 4{5}.0
fn roundInteger intg = if intg > 4 then 10 else 0
what is the name of the function? roundFloat
roundFloat 45.0 pre:10
–40.0
roundFloat 45.1 pre:10
–50.0
