OpenEuphoria

2. Re: the rad function

• Posted by Jeffrey Fielding <JJProg at CYBERBURY.NET> Sep 01, 1998
• 778 views
• Last edited Sep 02, 1998

John wrote:
>Roy Shepherd wrote:
>> John Worthinhton wrote
>> >> Roy Shepherd wrote:
> > >> Doug Cox wrote:

>Well, you get the idea.

>> >> function rad(atom sum)
>> >> sum =3D sum + (180/PI)
> > >> return sum
>> >> end function
>> >but .....  Shouldn't that be "sum =3D sum * 180 / PI" instead of +???
>> I stand corrected. Apologises to all. Well spotted John.

>All I can say is "WHEW".  ;')

>Also, seems like people are trying in the interest of speed to leave out
steps, so >going by that logic, would this be a hair better?  (That
sentence should be taken >out and shot, I think, but hopefully, it managed
to make sense anyway?)

>function rad2(atom sum)
>return sum*180/PI
>end function
>___
>_|ohn

Well, if you want even more speed, you could use
constant PI =3D  3.14159265359
constant PI_DIV_180 =3D 180 / PI
function rad(atom s)
        s =3D s * PI_DIV_180
        return s
end function

I wrote a simple testing program to test the following functions for rad:

constant PI =3D  3.14159265359
constant A =3D 180 / PI
function rad1(atom s)
    return s + (180 / PI)
end function
function rad2(atom s)
    return s * (180 / PI)
end function
function rad3(atom s)
    s =3D s + (180 / PI)
    return s
end function
function rad4(atom s)
    s =3D s * (180 / PI)
    return s
end function
function rad5(atom s)
    return s + A
end function
function rad6(atom s)
    return s * A
end function
function rad7(atom s)
    s =3D s + A
    return s
end function
function rad8(atom s)
    s =3D s * A
    return s
end function
sequence times, r
atom tm, a
r =3D {}
times =3D {}
for i =3D 1 to 8 do
    r =3D r & routine_id(sprintf("rad%d",i))
    times =3D times & 0
end for
-- You may want to change the maximim value for i or k because it runs kind
of
-- slowly.
for i =3D 1 to 100 do
    for j =3D 1 to 8 do
        tm =3D time()
        for k =3D 1 to 1000 do
            a =3D call_func(r[j],{i})
        end for
        tm =3D time() - tm
        if times[j] =3D 0 then
            times[j] =3D tm
        else
            times[j] =3D (times[j] + tm) / 2
        end if
    end for
end for
for i =3D 1 to 8 do
    printf(1, "%f\n",{times[i]})
end for

Jeffrey Fielding
JJProg at cyberbury.net