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-1

I have a 32-bit value that is stored in the VB.Net type Integer (i.e. Int32.) I am only interested in the bits - not the numerical value. Sometimes the 32nd bit is a one which is interpreted as a negative number. My goal is to reverse the actual bits. My original data is encoded into bits right-to-left (LSB right-most) and is read back in left-to-right (MSB left-most.) I am adapting someone else's code and design. One thought I had was maybe to convert to a long temporarily but I don't know how to do that and preserve the 32nd bit correctly.

Public Shared Function ReverseBits32(ByVal n As Integer) As Integer
 Dim result As Integer = 0

 For i As Integer = 0 To 32 - 1
 result = result * 2 + n Mod 2

 n = n >> 1 'n Or 2
 Next

 Return result
End Function

vb.net types type-conversion

share|improve this question

edited Mar 25 at 18:18

user339468

asked Mar 25 at 14:48

user339468user339468

10133 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  If you are interested in just the bit, why are you using signed integer?
  
  – the_lotus
  Mar 25 at 14:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I started with a sample application that originally worked with 8-bit values but was implemented using 32-bit integers. That makes sense as that will have much higher performance. My use requires 32-bit values. It would greatly delay my project if I tried to convert the data types at this point. I'll edit the question to more precisely explain my requirement.
  
  – user339468
  Mar 25 at 14:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It won't let me edit. So I need a function to completely reverse the bits in a 32 bit signed integer. I'm tripping over the 32nd bit when it is a one, thus making it a negative number.
  
  – user339468
  Mar 25 at 14:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So what you actually need is just a way to reverse the bits in a 32 bit integer, and that's what you're having trouble with? You don't necessarily need to go to 64 bits at all, as long as there's some other way of reversing the 32 bits?
  
  – Steven Doggart
  Mar 25 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, sorry I wasn't clearer.
  
  – user339468
  Mar 25 at 15:10
  

  
  
  
  

 | 
show 3 more comments

-1

I have a 32-bit value that is stored in the VB.Net type Integer (i.e. Int32.) I am only interested in the bits - not the numerical value. Sometimes the 32nd bit is a one which is interpreted as a negative number. My goal is to reverse the actual bits. My original data is encoded into bits right-to-left (LSB right-most) and is read back in left-to-right (MSB left-most.) I am adapting someone else's code and design. One thought I had was maybe to convert to a long temporarily but I don't know how to do that and preserve the 32nd bit correctly.

Public Shared Function ReverseBits32(ByVal n As Integer) As Integer
 Dim result As Integer = 0

 For i As Integer = 0 To 32 - 1
 result = result * 2 + n Mod 2

 n = n >> 1 'n Or 2
 Next

 Return result
End Function

vb.net types type-conversion

share|improve this question

edited Mar 25 at 18:18

user339468

asked Mar 25 at 14:48

user339468user339468

10133 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  If you are interested in just the bit, why are you using signed integer?
  
  – the_lotus
  Mar 25 at 14:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  I started with a sample application that originally worked with 8-bit values but was implemented using 32-bit integers. That makes sense as that will have much higher performance. My use requires 32-bit values. It would greatly delay my project if I tried to convert the data types at this point. I'll edit the question to more precisely explain my requirement.
  
  – user339468
  Mar 25 at 14:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It won't let me edit. So I need a function to completely reverse the bits in a 32 bit signed integer. I'm tripping over the 32nd bit when it is a one, thus making it a negative number.
  
  – user339468
  Mar 25 at 14:57
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  So what you actually need is just a way to reverse the bits in a 32 bit integer, and that's what you're having trouble with? You don't necessarily need to go to 64 bits at all, as long as there's some other way of reversing the 32 bits?
  
  – Steven Doggart
  Mar 25 at 15:06
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Yes, sorry I wasn't clearer.
  
  – user339468
  Mar 25 at 15:10
  

  
  
  
  

 | 
show 3 more comments

I have a 32-bit value that is stored in the VB.Net type Integer (i.e. Int32.) I am only interested in the bits - not the numerical value. Sometimes the 32nd bit is a one which is interpreted as a negative number. My goal is to reverse the actual bits. My original data is encoded into bits right-to-left (LSB right-most) and is read back in left-to-right (MSB left-most.) I am adapting someone else's code and design. One thought I had was maybe to convert to a long temporarily but I don't know how to do that and preserve the 32nd bit correctly.

Public Shared Function ReverseBits32(ByVal n As Integer) As Integer
 Dim result As Integer = 0

 For i As Integer = 0 To 32 - 1
 result = result * 2 + n Mod 2

 n = n >> 1 'n Or 2
 Next

 Return result
End Function

vb.net types type-conversion

share|improve this question

edited Mar 25 at 18:18

user339468

asked Mar 25 at 14:48

user339468user339468

10133 bronze badges

Public Shared Function ReverseBits32(ByVal n As Integer) As Integer
 Dim result As Integer = 0

 For i As Integer = 0 To 32 - 1
 result = result * 2 + n Mod 2

 n = n >> 1 'n Or 2
 Next

 Return result
End Function

share|improve this question

edited Mar 25 at 18:18

user339468

asked Mar 25 at 14:48

user339468user339468

10133 bronze badges

share|improve this question

edited Mar 25 at 18:18

user339468

asked Mar 25 at 14:48

user339468user339468

10133 bronze badges

asked Mar 25 at 14:48

user339468user339468

10133 bronze badges

asked Mar 25 at 14:48

user339468user339468

10133 bronze badges

1 Answer
1

active

oldest

votes

1

If you had a method to reverse the bits of a byte you could apply it four times to the bytes of an integer. A little research finds Bit Twiddling Hacks.

Module Module1

 Sub ShowBits(a As Integer)
 Dim aa = BitConverter.GetBytes(a)
 Console.WriteLine(String.Join(" ", aa.Select(Function(b) Convert.ToString(b, 2).PadLeft(8, "0"c))))
 End Sub

 Function ReverseBits(b As Byte) As Byte
 ' From https://graphics.stanford.edu/~seander/bithacks.html#ReverseByteWith32Bits
 Dim c = CULng(b)
 Return CByte((((c * &H802UL And &H22110UL) Or (c * &H8020UL And &H88440UL)) * &H10101UL >> 16) And &HFFUL)
 End Function

 Function ReverseBits(a As Integer) As Integer
 Dim bb = BitConverter.GetBytes(a)
 Dim cc(3) As Byte
 For i = 0 To 3
 cc(3 - i) = ReverseBits(bb(i))
 Next

 Return BitConverter.ToInt32(cc, 0)

 End Function

 Sub Main()

 Dim y = -762334566
 ShowBits(y)
 y = ReverseBits(y)
 ShowBits(y)

 Console.ReadLine()

 End Sub

End Module

Output from test value:

10011010 10110010 10001111 11010010
01001011 11110001 01001101 01011001

I used the "no 64-bit" method because it is written for a language where arithmetic overflow is ignored - the methods using 64-bit operations rely on that but it is not the default for VB.NET.

share|improve this answer

answered Mar 25 at 18:26

Andrew MortonAndrew Morton

16.5k66 gold badges3838 silver badges5454 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This will take me some time to puzzle out but I assume "CByte((((c * &H802UL And &H22110UL) Or (c * &H8020UL And &H88440UL)) * &H10101UL >> 16) And &HFFUL)" is handling the sign bit? This looks promising.
  
  – user339468
  Mar 25 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  A Byte is unsigned. BitConverter.GetBytes gets the raw bytes without any interpretation. So the sign bit does not come into it :)
  
  – Andrew Morton
  Mar 25 at 19:05
  

  
  
  
  

add a comment | 

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1

If you had a method to reverse the bits of a byte you could apply it four times to the bytes of an integer. A little research finds Bit Twiddling Hacks.

Module Module1

 Sub ShowBits(a As Integer)
 Dim aa = BitConverter.GetBytes(a)
 Console.WriteLine(String.Join(" ", aa.Select(Function(b) Convert.ToString(b, 2).PadLeft(8, "0"c))))
 End Sub

 Function ReverseBits(b As Byte) As Byte
 ' From https://graphics.stanford.edu/~seander/bithacks.html#ReverseByteWith32Bits
 Dim c = CULng(b)
 Return CByte((((c * &H802UL And &H22110UL) Or (c * &H8020UL And &H88440UL)) * &H10101UL >> 16) And &HFFUL)
 End Function

 Function ReverseBits(a As Integer) As Integer
 Dim bb = BitConverter.GetBytes(a)
 Dim cc(3) As Byte
 For i = 0 To 3
 cc(3 - i) = ReverseBits(bb(i))
 Next

 Return BitConverter.ToInt32(cc, 0)

 End Function

 Sub Main()

 Dim y = -762334566
 ShowBits(y)
 y = ReverseBits(y)
 ShowBits(y)

 Console.ReadLine()

 End Sub

End Module

Output from test value:

10011010 10110010 10001111 11010010
01001011 11110001 01001101 01011001

I used the "no 64-bit" method because it is written for a language where arithmetic overflow is ignored - the methods using 64-bit operations rely on that but it is not the default for VB.NET.

share|improve this answer

answered Mar 25 at 18:26

Andrew MortonAndrew Morton

16.5k66 gold badges3838 silver badges5454 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This will take me some time to puzzle out but I assume "CByte((((c * &H802UL And &H22110UL) Or (c * &H8020UL And &H88440UL)) * &H10101UL >> 16) And &HFFUL)" is handling the sign bit? This looks promising.
  
  – user339468
  Mar 25 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  A Byte is unsigned. BitConverter.GetBytes gets the raw bytes without any interpretation. So the sign bit does not come into it :)
  
  – Andrew Morton
  Mar 25 at 19:05
  

  
  
  
  

add a comment | 

1

If you had a method to reverse the bits of a byte you could apply it four times to the bytes of an integer. A little research finds Bit Twiddling Hacks.

Module Module1

 Sub ShowBits(a As Integer)
 Dim aa = BitConverter.GetBytes(a)
 Console.WriteLine(String.Join(" ", aa.Select(Function(b) Convert.ToString(b, 2).PadLeft(8, "0"c))))
 End Sub

 Function ReverseBits(b As Byte) As Byte
 ' From https://graphics.stanford.edu/~seander/bithacks.html#ReverseByteWith32Bits
 Dim c = CULng(b)
 Return CByte((((c * &H802UL And &H22110UL) Or (c * &H8020UL And &H88440UL)) * &H10101UL >> 16) And &HFFUL)
 End Function

 Function ReverseBits(a As Integer) As Integer
 Dim bb = BitConverter.GetBytes(a)
 Dim cc(3) As Byte
 For i = 0 To 3
 cc(3 - i) = ReverseBits(bb(i))
 Next

 Return BitConverter.ToInt32(cc, 0)

 End Function

 Sub Main()

 Dim y = -762334566
 ShowBits(y)
 y = ReverseBits(y)
 ShowBits(y)

 Console.ReadLine()

 End Sub

End Module

Output from test value:

10011010 10110010 10001111 11010010
01001011 11110001 01001101 01011001

I used the "no 64-bit" method because it is written for a language where arithmetic overflow is ignored - the methods using 64-bit operations rely on that but it is not the default for VB.NET.

share|improve this answer

answered Mar 25 at 18:26

Andrew MortonAndrew Morton

16.5k66 gold badges3838 silver badges5454 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  This will take me some time to puzzle out but I assume "CByte((((c * &H802UL And &H22110UL) Or (c * &H8020UL And &H88440UL)) * &H10101UL >> 16) And &HFFUL)" is handling the sign bit? This looks promising.
  
  – user339468
  Mar 25 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  A Byte is unsigned. BitConverter.GetBytes gets the raw bytes without any interpretation. So the sign bit does not come into it :)
  
  – Andrew Morton
  Mar 25 at 19:05
  

  
  
  
  

add a comment | 

If you had a method to reverse the bits of a byte you could apply it four times to the bytes of an integer. A little research finds Bit Twiddling Hacks.

Module Module1

 Sub ShowBits(a As Integer)
 Dim aa = BitConverter.GetBytes(a)
 Console.WriteLine(String.Join(" ", aa.Select(Function(b) Convert.ToString(b, 2).PadLeft(8, "0"c))))
 End Sub

 Function ReverseBits(b As Byte) As Byte
 ' From https://graphics.stanford.edu/~seander/bithacks.html#ReverseByteWith32Bits
 Dim c = CULng(b)
 Return CByte((((c * &H802UL And &H22110UL) Or (c * &H8020UL And &H88440UL)) * &H10101UL >> 16) And &HFFUL)
 End Function

 Function ReverseBits(a As Integer) As Integer
 Dim bb = BitConverter.GetBytes(a)
 Dim cc(3) As Byte
 For i = 0 To 3
 cc(3 - i) = ReverseBits(bb(i))
 Next

 Return BitConverter.ToInt32(cc, 0)

 End Function

 Sub Main()

 Dim y = -762334566
 ShowBits(y)
 y = ReverseBits(y)
 ShowBits(y)

 Console.ReadLine()

 End Sub

End Module

Output from test value:

10011010 10110010 10001111 11010010
01001011 11110001 01001101 01011001

I used the "no 64-bit" method because it is written for a language where arithmetic overflow is ignored - the methods using 64-bit operations rely on that but it is not the default for VB.NET.

share|improve this answer

answered Mar 25 at 18:26

Andrew MortonAndrew Morton

16.5k66 gold badges3838 silver badges5454 bronze badges

Module Module1

 Sub ShowBits(a As Integer)
 Dim aa = BitConverter.GetBytes(a)
 Console.WriteLine(String.Join(" ", aa.Select(Function(b) Convert.ToString(b, 2).PadLeft(8, "0"c))))
 End Sub

 Function ReverseBits(b As Byte) As Byte
 ' From https://graphics.stanford.edu/~seander/bithacks.html#ReverseByteWith32Bits
 Dim c = CULng(b)
 Return CByte((((c * &H802UL And &H22110UL) Or (c * &H8020UL And &H88440UL)) * &H10101UL >> 16) And &HFFUL)
 End Function

 Function ReverseBits(a As Integer) As Integer
 Dim bb = BitConverter.GetBytes(a)
 Dim cc(3) As Byte
 For i = 0 To 3
 cc(3 - i) = ReverseBits(bb(i))
 Next

 Return BitConverter.ToInt32(cc, 0)

 End Function

 Sub Main()

 Dim y = -762334566
 ShowBits(y)
 y = ReverseBits(y)
 ShowBits(y)

 Console.ReadLine()

 End Sub

End Module

10011010 10110010 10001111 11010010
01001011 11110001 01001101 01011001

share|improve this answer

answered Mar 25 at 18:26

Andrew MortonAndrew Morton

16.5k66 gold badges3838 silver badges5454 bronze badges

answered Mar 25 at 18:26

Andrew MortonAndrew Morton

16.5k66 gold badges3838 silver badges5454 bronze badges

answered Mar 25 at 18:26

Andrew MortonAndrew Morton

16.5k66 gold badges3838 silver badges5454 bronze badges