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Answers for "excelvba 32-bit long integer to bits"

VBA
14

excelvba 32-bit long integer to bits

Function LongToBits$(ByVal n&)
    Dim i&
    LongToBits = "00000000000000000000000000000000"
    If n And &H80000000 Then
        Mid$(LongToBits, 1, 1) = "1"
        n = n And Not &H80000000
    End If
    For i = 32 To 2 Step -1
        If n And 1 Then Mid$(LongToBits, i, 1) = "1"
        n = n \ 2
    Next
End Function

'------------------------------------------------------------------------------

MsgBox ByteToBits(0)			'<--displays: 00000000000000000000000000000000
MsgBox LongToBits(293781237)	'<--displays: 00010001100000101011111011110101
MsgBox ByteToBits(-1)			'<--displays: 11111111111111111111111111111111
Posted by: Guest on May-23-2020
26

excelvba is bit set in 64-bit integer

'Extremely fast VBA function to test if a specified bit is set 
'within a 64-bit LongLong integer.

'Bits are numbered from 0 to 63, so the first (least significant) bit
'is bit 0.

'Note: we do not inspect bit 63, the sign bit.
'Note: the LongLong data type is only available in 64-bit VBA.


Function LongLongBitIsSet(theLongLong^, bit As Byte) As Boolean
    Static i&, b^()
    If (Not Not b) = 0 Then
        ReDim b(0 To 62)
        For i = 0 To 62
            b(i) = 2 ^ i
        Next
    End If
    If bit < 63 Then LongLongBitIsSet = theLongLong And b(bit)
End Function


'------------------------------------------------------------------
MsgBox LongBitIsSet(255, 7)                       '<--displays: True
MsgBox LongBitIsSet(230, 0)                       '<--displays: False
MsgBox LongBitIsSet(16384, 14)                    '<--displays: True
MsgBox LongBitIsSet(-16383, 0)                    '<--displays: True
MsgBox LongBitIsSet(&H7FFFFFFF, 0), 0)            '<--displays: True
MsgBox LongLongBitIsSet(&H7FFFFFFFFFFFFFff^, 62)  '<--displays: True
'
'
'
Posted by: Guest on February-05-2021
20

vba is specific bit set in long integer

'Extremely fast VBA function to test if a specified bit is set 
'within a 32-bit Long integer.

'Bits are numbered from 0 to 31, so the first (least significant) bit
'is bit 0.

'Note: we do not inspect bit 31, the sign bit.


Function LongBitIsSet(theLong&, bit As Byte) As Boolean
    Static i&, b&()
    If (Not Not b) = 0 Then
        ReDim b(0 To 30)
        For i = 0 To 30
            b(i) = 2 ^ i
        Next
    End If
    If bit < 31 Then LongBitIsSet = theLong And b(bit)
End Function


'------------------------------------------------------------------
MsgBox LongBitIsSet(255, 7)             '<--displays: True
MsgBox LongBitIsSet(230, 0)             '<--displays: False
MsgBox LongBitIsSet(16384, 14)          '<--displays: True
MsgBox LongBitIsSet(-16383, 0)          '<--displays: True
MsgBox LongBitIsSet(&H7FFFFFFF, 0)      '<--displays: True
Posted by: Guest on February-05-2021

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