Check Bitwise flags

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A

Anil Gupte

This is somethings I used to do often in other languages (such as Lisp), but
don't know how to do in VB.

I want to check if a certain bit (flag) is set. For example if I want to
check whether the 2nd bit is set, I compare it to binary 2

7 and 2 should return true (111 AND 010 = true)

How does one do this in VB?

Thanx,
 
Anil said:
This is somethings I used to do often in other languages (such as Lisp), but
don't know how to do in VB.

I want to check if a certain bit (flag) is set. For example if I want to
check whether the 2nd bit is set, I compare it to binary 2

7 and 2 should return true (111 AND 010 = true)

How does one do this in VB?
Click to expand...

The same way,

Sub Main()
Dim MyByte As Byte = &HAA
Dim I As Integer
WriteLine("Bit_no :____8____7____6____5____4____3____2____1")
Write("Bitval :")
For I = 8 To 1 Step -1
Write((2 ^ (I - 1)).ToString.PadLeft(5))
Next
WriteLine()
Write("current:")
For I = 8 To 1 Step -1
If MyByte And (2 ^ (I - 1)) Then
Write((2 ^ (I - 1)).ToString.PadLeft(5))
Else
Write(" 0")
End If
Next
Write(MyByte.ToString.PadLeft(5))
ReadLine()
End Sub

HTH
Matthias
 
Anil Gupte said:
This is somethings I used to do often in other languages (such as Lisp),
but don't know how to do in VB.

I want to check if a certain bit (flag) is set. For example if I want to
check whether the 2nd bit is set, I compare it to binary 2

7 and 2 should return true (111 AND 010 = true)

How does one do this in VB?
Click to expand...

You already gave the answer by yourself. I don't quite see why you need a
function to do this, but who cares. So:


Function CheckBitMask(Bitmask As Long, Value As Long) As Boolean
Return CBool(Bitmask And Value)
End Function

Sub Foo()
'> 7 and 2 should return true (111 AND 010 = true)
Console.WriteLine(CheckBitMask(2, 7).ToString)
End Sub
 
Anil Gupte said:
This is somethings I used to do often in other languages (such as Lisp),
but don't know how to do in VB.

I want to check if a certain bit (flag) is set. For example if I want to
check whether the 2nd bit is set, I compare it to binary 2

7 and 2 should return true (111 AND 010 = true)
Click to expand...

\\\
If CBool(Foo And &H2) Then
...
End If
///
 
Thanx, I will check out both solutions (yours and Heikki's).

Of course Herfrieds solution is the way to go, since there is no need to
wrap a logical operation such as foo And bar into a function. I realized
later that you didn't even ask for a function, just a solution. Sometimes a
solution is far too obvious to be seen directly.

-h-
 
Herfried K. Wagner said:
\\\
If CBool(Foo And &H2) Then
...
End If
///
Click to expand...

Should it not be:

If Foo And &H2 = &H2
...
End If

?? Performing a bitwise AND will give a result of &H2 if &H2 is set to ON
and will give you 0 if &H2 is not set... Also, doing it this way, you don't
need to convert to boolean since adding "= &H2" makes it a boolean
expression...

HTH,
Mythran
 
Should it not be:
If Foo And &H2 = &H2
Click to expand...
?? Performing a bitwise AND will give a result of &H2 if &H2 is set to ON
and will give you 0 if &H2 is not set... Also, doing it this way, you
don't need to convert to boolean since adding "= &H2" makes it a boolean
expression...
Click to expand...

What else could Foo And &H2 be than 0 and &h2? I don't quite see your point,
there is only one bit set in &H2. In languages such as c/c++ there is no
need to cast something to boolean, since any nonzero value is defined to be
"true".

-h-
 
Heikki Leivo said:
What else could Foo And &H2 be than 0 and &h2? I don't quite see your
point, there is only one bit set in &H2.
Click to expand...

'CBool' will return 'True' if a bit in the value passed to it is set to 1.
In languages such as c/c++ there is no need to cast something to boolean,
since any nonzero value is defined to be "true".
Click to expand...

Such automatic conversions are performed with 'Option Strict Off', but I
think that the more explicit way is the better choice
most times.
 
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