How do I split a 16-bit value into two 8-bit values in VB.NET?

Question

I have a 16-bit value like this:

 0000000000000011

I want to split this in two 8-bit values like:

 00000000    and    00000011

The 16-bit value is a variable, byte_val1, which is a unsigned 16-bit integer.

How can I do this in VB.NET?

Answer 1

You can use the BitConverter class:

Dim bytes As Byte() = BitConverter.GetBytes(byte_val1)

Now bytes(0) contains the lower byte (00000011) and bytes(1) contains the higher byte (00000000).

Answer 2

To obtain lower 8 bits, you need to and it with 11111111, or 0xff
To obtain upper 8 bits, you need to and it with 1111111100000000, or 0xff00
Then to an arithmetic left shift exactly 8 bits.

I have not done Visual Basic in a while, but the idea is as follows:

word a = 1930392;
byte alower = a & 0xff;
byte ahigher = (a & 0xff00) >> 8;

Oh, here's a code I made with three textboxes named a, alower, aupper and a button btnDoIt:

Private Sub btnDoIt_Click(sender As Object, e As EventArgs) Handles btnDoIt.Click
    Dim localA = Integer.Parse(a.Text())
    Dim localALower = localA And &HFF
    Dim localAUpper = (localA And &HFF00) >> 8

    alower.Text = localALower
    aupper.Text = localAUpper
End Sub

Sidenote: (localA And &HFF00) >> 8 is equivelent to (localA >> 8) And &HFF

Answer 3

A simple bitshift should work if you want certain byte(s) from any value (32 bit, 64 bit, 128 bit, etc..).

The simplest way is to AND ("binary AND", also written as "&") the source with the bits you want and then to shift it as many times as needed to get the desired value.

Dim initial as Integer = &H123456 ' Hex: 0x123456
Dim result as Byte = (initial AND &H00FF00) >> 8
'result = 0x34

and if you would like to join them again it's even simpler:

Dim initial as Integer = &H123456 ' Hex: 0x123456
Dim result_a as Byte = (initial AND &H00FF00) >> 8
Dim result_b as Byte = (initial AND &HFF0000) >> 16
Dim final as Integer = (result_a << 8) or (result_b << 16)
'final = 0x123400