Most of you who have tried to accomplish this, know that Visual Basic doesn't have any function for bit shifting. And bit shifting si very usefull
Some of you probably know how to shift bits, but this tutorial is for those who don't know how to do it...Shifting bits is actualy easy, we use multiplication and division by powers of 2.
For right shifting, we use division:
CODE
NewValue = Value \ (2 ^ BitsToShift)
And for left shifting, we use multiplication:
CODE
NewValue = Value * (2 ^ BitsToShift)
Now, that's not all... With Visual Basic, things can get a little trickier
Trouble with VB is that it's Integer (16bit), and Long(32bit) data types are signed, and what that means they can hold both positive and negative values. One bit is used to hold the sign. All remaining bits are used to store actual value.
Integer data type can store 65.535 numbers, but, since it is signed, it can only store numbers from -32.768 upto 32.767.
Take a lok at this (for all you visual types, like me
)CODE
DEC BIN HEX OCT
1 0000000000000001 &H1 1
2 0000000000000010 &H2 2
32.766 0111111111111110 &H7FFE 77.776
32.767 0111111111111111 &H7FFF 77.777
-32.767 1000000000000000 &H8000 100.000
-32.766 1000000000000001 &H8001 100.001
-2 1111111111111110 &HFFFE 177.776
-1 1111111111111111 &HFFFF 177.777
If you take the binary value of -1 (1111111111111111), and convert it to decimal in calculator, you wil get a value of 65.535. Same goes for hexadecimal and octal values... Neat, right
Now, what does this have to do with bit shifting? It has everything to do, because during bit shifting to the right, sign bit is preserved in Visual Basic, and that gives wrong values in the end... Now, this explained, we can go on, and write that shifting function...
SHIFTING BITS RIGHT
In most cases, when people want to shift bits, they want to use unsigned shifting, that is, no negative values. When such a shift occurs, shift bit is not preserved. Now, this requires one extra step in order to work with Visual Basic, and it's signed data types. When value is negative, you need to AND it. Here's the example code
CODE
If Value < 0 Then
NewValue = ((Value And &HFFFE) \ 2) And &H7FFF ' * Shift 1 bit right
' NewValue = ((Value And &HFFFC) \ 4) And &H3FFF ' * Shift 2 bits right
' NewValue = ((Value And &HFFF8) \ 8) And &H1FFF ' * Shift 3 bits right
' NewValue = ((Value And &HFFF0) \ 16) And &HEFFF ' * Shift 4 bits right
' NewValue = ((Value And &HFFE0) \ 32) And &HF7FF ' * Shift 5 bits right
' NewValue = ((Value And &HFFC0) \ 64) And &HFBFF ' * Shift 6 bits right
' NewValue = ((Value And &HFF80) \ 128) And &HFDFF ' * Shift 7 bits right
' NewValue = ((Value And &HFF00) \ 256) And &HFEFF ' * Shift 8 bits right
Else
NewValue = Value \ (2 ^ BitsToShift)
End If
SHIFTING BITS LEFT
Shifting bits to the left, doesn't require any additional operations, it is simply multiplied by the power of 2, and is same for the signed and unsigned numbers:
CODE
NewValue = Value * (2 ^ BitsToShift)
Some people recommend using Long data type, for Integer Values, in order to use unsigned bit shifting. Others, turning off 'Integer Overflow Check' in 'Advanced Optimisations'... Unsigned right bit shifting can't be tested from the Visual Basic IDE, instead, compile it into a project, turn off already mentined optimisation, and try it that way...
That's all folks
