Bit wise option code?

I think that's what it's called. I've seen where you can do something like this.

Const GLOW = 2
Const FOG = 4
Const SKY = 8
Const GROUND = 16

MakeWorld(GLOW+FOG+GROUND)

Then the function can somehow figure out what options you pick. Anyone know how to code this?

If flags and GLOW

;glow enabled

EndIf

If flags and FOG

;FOG enabled

EndIf

etc..

Const Glow = 2
Const Rings = 4
Const LensRef = 6

PrintOptions(Glow+Rings)

Function PrintOptions(Flag)

If Flag And Glow Then Print "Glow"
If Flag And Rings Then Print "Ring"
If Flag And LensRef Then Print "LensRef"

End Function

WaitKey()


I tried this, it didn't seem to work.

Nevermid, 6 had to be an 8

Using these bit-flags are handy for setting a set of true/false values in one integer, instead of using a variable for every true/false value you have. Basically there are 32-bits in a number and each bit can either be set to 0 (false) or 1 (true).

I always use the shift left operation to set up a bunch of flags - it's easier to follow, something like this:

Const kF_Glow     = 1 Shl 0
Const kF_Rings	  = 1 Shl 1
Const kF_LensRef  = 1 Shl 2

You set flags like this:

Flags = kF_Glow Or kF_Rings Or kF_LensRef

Or like this:

Flags = Flags Or kF_Rings

(you can also use + instead of 'or' if you like, but 'or' is safer because if the flag is already set it won't cause a problem. If you use + and the flag is already set it will break because it is adding to the number, so you'll get a different value)

You remove flags like this:

Flags = Flags And ~kF_Glow

You can test if a flag is set like this:

If (Flags And kF_Glow) Then do something...

You can test if a flag isn't set like this:

If ( (Flags And kF_Glow) = 0 ) Then do even more stuff...

You can even test if a bunch of flags are set like this:

CheckFlags = kF_Glow Or kF_Rings
if ((Flags And CheckFlags) = CheckFlags) ) Then do stuff...

hope that helps,

Here is a further variation, maybe useful to see what happens:

If (Flag And Glow) = Glow Then Print "Glow"

Let's say flag is 7 and Glow is 4

"(Flag AND GLow)" is an expression that forces a calculation. It will now set those bits in the result which are true in both, in Flag AND in Glow.

Flag: binary 0111
Glow: binary 0100

You see there is only one bit set in both numbers, the third bit. So the result is 0100

So
(%0111 AND %0100) = %0100
or we could say:
(7 and 4) = 4

If the third bit in Flag would not be set, then the result would be zero:
(3 and 4)=0
or
(0011 AND 0100) = 0000
because none of the bits are set in both variables.