Bits of a byte?

BlitzMax Forums/BlitzMax Beginners Area/Bits of a byte?

Ked	(Posted 2009) [#1]

How would I get the individual 8 bits from a byte? Like:

bit1=howeveryoudoit
bit2=howeveryoudoit
bit3=howeveryoudoit
bit4=howeveryoudoit
etc...

Jesse

(Posted 2009) [#2]

bit1 = n & %1
bit2 = n & %10
bit3 = n & %100
bit4 = n & %1000

Ked	(Posted 2009) [#3]

That's a safe way of doing it? No Shr or Shl?

Jesse

(Posted 2009) [#4]

true. but there are going to be times when you'll need to use the shift.

Ked	(Posted 2009) [#5]

... Then, what's an all-the-time-good way of retrieving the bits in a byte?

SebHoll

(Posted 2009) [#6]

bit1 = (n Shr 0) & %1
bit2 = (n Shr 1) & %1
bit3 = (n Shr 2) & %1
bit4 = (n Shr 3) & %1

Obviously, the bit indices themselves depend on how you're numbering your bits (from LSB->MSB in this case) and maybe even the endian-ness of the processor architecture. But to all intents and purposes, this should be what you are after. :-)

Ked	(Posted 2009) [#7]

Great! And I could always slip in a ?LittleEndian if needed. :) Thanks!

Jesse

(Posted 2009) [#8]

SebHoll's way is correct and usefull if that is what you need. but it is not allways necesary to shift. With the speed of modern processors you probably would not notice any difference with the shift and with out it anyway. I sometimes do something like this:

const lft:byte = %100
const cent:byte = %010
const rght:byte = %001
.
.
.
gunsInSHip = lft|rght
.
.
.
if (gunsInShip & lft) then shoot left gun
if (gunsInShip & rgt) then shoot right gun
if (gunsInSHip & cent) then shoot center gun

obviously the ship will shoot the left and the right gun.
this works with out having to shift any bits.
and logically this way is faster in this case.

Shortwind

(Posted 2009) [#9]

Some Old Timey Pre-School Methods....

SuperStrict

Local n:Byte, bit7:Byte, bit6:Byte, bit5:Byte, bit4:Byte, bit3:Byte, bit2:Byte, bit1:Byte, bit0:Byte
Local bitty:Byte[8]

n=195

'Old School Method #1 - High Bit -> Low Bit (Normal Representation)
bit7=(n/128) Mod 2
bit6=(n/64) Mod 2
bit5=(n/32) Mod 2
bit4=(n/16) Mod 2
bit3=(n/8) Mod 2
bit2=(n/4) Mod 2
bit1=(n/2) Mod 2
bit0=(n/1) Mod 2

Print "Byte value -> "+n+" = "+bit7+bit6+bit5+bit4+bit3+bit2+bit1+bit0+" bits"
Print "~n"

'Old School Method #2 - High Bit -> Low Bit (Normal Representation) (Good for array)
Local temp:Byte=n
For Local i:Int = 7 To 0 Step -1
	If 2^i <= temp Then
		bitty[i]=1
		temp=temp-(2^i)
	Else
		bitty[i]=0
	End If
Next

Print "Byte Value -> "+n+" = "
For Local i:Int = 7 To 0 Step -1
	Print bitty[i]
Next
Print "bits."

:()

Shortwind

(Posted 2009) [#10]

Here are some usefull functions found right in the code archives.

This is from S|F....

Const BIT_0			 = 1
Const BIT_1			 = 2
Const BIT_2			 = 4
Const BIT_3			 = 8
Const BIT_4			 = 16
Const BIT_5			 = 32
Const BIT_6			 = 64
Const BIT_7			 = 128

Function SetBit(B:Byte Var, Bit:Int) 
	B :| Bit
End Function

Function ClearBit(B:Byte Var, Bit:Int) 
	B :- Bit
End Function

Function GetBit:Int(B:Byte, Bit:Int)
	Return B & Bit > 0
End Function

' - TEST CODE, DELETE THIS.
Global Test:Byte = 0

SetBit(Test, BIT_0) 
SetBit(Test, BIT_2) 
SetBit(Test, BIT_4) 
SetBit(Test, BIT_6) 
ClearBit(Test, BIT_6)
Print("BIT_0: " + GetBit(Test, BIT_0) ) 
Print("BIT_1: " + GetBit(Test, BIT_1))
Print("BIT_2: " + GetBit(Test, BIT_2))
Print("BIT_3: " + GetBit(Test, BIT_3))
Print("BIT_4: " + GetBit(Test, BIT_4) ) 
Print("BIT_5: " + GetBit(Test, BIT_5))
Print("BIT_6: " + GetBit(Test, BIT_6))
Print("BIT_7: " + GetBit(Test, BIT_7))