Xl Vba Test If Specific Bits Set In Integer

[Solved] Xl Vba Test If Specific Bits Set In Integer | Vb - Code Explorer | yomemimo.com
Question : vba test if particular bits set in longlong integer

Answered by : daniel-ferry

'Fast VBA function to test if specific bits are set
'within a 64-bit LongLong integer. If ALL specified bits are set then
'this function returns True. If ANY of the specified bits are
'not set then this function returns False.
'Bits are numbered from 0 to 63, so the first (least significant) bit
'is bit 0.
'Note: bits can be specified in any order.
'Note: any number of bits can be specified in one call.
'Note: the LongLong data type is only available in 64-bit VBA.
Function LongLongBitsAreSet(theLongLong^, ParamArray bits()) As Boolean Static i&, b^() If UBound(bits) = -1 Then Exit Function If (Not Not b) = 0 Then ReDim b(0 To 62) For i = 0 To 62 b(i) = 2 ^ i Next End If For i = 0 To UBound(bits) If bits(i) < 0 Then Exit Function If bits(i) > 62 Then Exit Function If (theLongLong And b(Int(bits(i)))) = 0 Then Exit Function Next LongLongBitsAreSet = True
End Function
'----------------------------------------------------------------------------------------------------------------------------------------------------------------------
MsgBox LongLongBitsAreSet(255, 7) '<--displays: True 255 = 00000000 00000000 00000000 00000000 00000000 00000000 00000000 11111111
MsgBox LongLongBitsAreSet(230, 0) '<--displays: False 230 = 00000000 00000000 00000000 00000000 00000000 00000000 00000000 11100110
MsgBox LongLongBitsAreSet(85, 0, 2, 6, 4) '<--displays: True 85 = 00000000 00000000 00000000 00000000 00000000 00000000 00000000 01010101
MsgBox LongLongBitsAreSet(85, 0, 2, 5, 4) '<--displays: False 85 = 00000000 00000000 00000000 00000000 00000000 00000000 00000000 01010101
MsgBox LongLongBitsAreSet(485, 2, 7, 5) '<--displays: True 485 = 00000000 00000000 00000000 00000000 00000000 00000000 00000001 11100101
MsgBox LongLongBitsAreSet(24585, 14, 13, 3) '<--displays: True 24585 = 00000000 00000000 00000000 00000000 00000000 00000000 01100000 00001001
MsgBox LongLongBitsAreSet(&H7D12EACB, 24, 26, 30, 7) '<--displays: True 2098391755 = 00000000 00000000 00000000 00000000 01111101 00010010 11101010 11001011
MsgBox LongLongBitsAreSet(2 ^ 62 + 1, 62, 0) '<--displays: True 4611686018427387905 = 01000000 00000000 00000000 00000000 00000000 00000000 00000000 00000001
'
'
'

Source : http://academy.excelhero.com/ | Last Update : Sun, 14 Feb 21

Question : xl vba check if specific bits are set in int32

Answered by : daniel-ferry

'Fast VBA function to test if specific bits are set
'within a 32-bit Long integer. If ALL specified bits are set then
'this function returns True. If ANY of the specified bits are
'not set then this function returns False.
'Bits are numbered from 0 to 31, so the first (least significant) bit
'is bit 0.
'Note: bits can be specified in any order.
'Note: any number of bits can be specified in one call.
Function LongBitsAreSet(theLong&, ParamArray bits()) As Boolean Static i&, b&() If UBound(bits) = -1 Then Exit Function If (Not Not b) = 0 Then ReDim b(0 To 30) For i = 0 To 30 b(i) = 2 ^ i Next End If For i = 0 To UBound(bits) If bits(i) < 0 Then Exit Function If bits(i) > 30 Then Exit Function If (theLong And b(Int(bits(i)))) = 0 Then Exit Function Next LongBitsAreSet = True
End Function
'---------------------------------------------------------------------------------------------------------------------
MsgBox LongBitsAreSet(255, 7) '<--displays: True 255 = 00000000 00000000 00000000 11111111
MsgBox LongBitsAreSet(230, 0) '<--displays: False 230 = 00000000 00000000 00000000 11100110
MsgBox LongBitsAreSet(85, 0, 2, 6, 4) '<--displays: True 85 = 00000000 00000000 00000000 01010101
MsgBox LongBitsAreSet(85, 0, 2, 5, 4) '<--displays: False 85 = 00000000 00000000 00000000 01010101
MsgBox LongBitsAreSet(485, 2, 7, 5) '<--displays: True 485 = 00000000 00000000 00000001 11100101
MsgBox LongBitsAreSet(24585, 14, 13, 3) '<--displays: True 24585 = 00000000 00000000 01100000 00001001
MsgBox LongBitsAreSet(&H7D12EACB, 24, 26, 30, 7) '<--displays: True 2098391755 = 01111101 00010010 11101010 11001011
'
'
'

Source : http://academy.excelhero.com/ | Last Update : Sun, 14 Feb 21

Question : vba test if specific bits set in integer

Answered by : daniel-ferry

'Fast VBA function to test if specific bits are set
'within a 16-bit Integer. If ALL specified bits are set then
'this function returns True. If ANY of the specified bits are
'not set then this function returns False.
'Bits are numbered from 0 to 15, so the first (least significant) bit
'is bit 0.
'Note: bits can be specified in any order.
'Note: any number of bits can be specified in one call.
Function IntegerBitsAreSet(theInteger%, ParamArray bits()) As Boolean Static i&, b%() If UBound(bits) = -1 Then Exit Function If (Not Not b) = 0 Then ReDim b(0 To 14) For i = 0 To 14 b(i) = 2 ^ i Next End If For i = 0 To UBound(bits) If bits(i) < 0 Then Exit Function If bits(i) > 14 Then Exit Function If (theInteger And b(Int(bits(i)))) = 0 Then Exit Function Next IntegerBitsAreSet = True
End Function
'----------------------------------------------------------------------------------------
MsgBox IntegerBitsAreSet(255, 7) '<--displays: True 255 = 00000000 11111111
MsgBox IntegerBitsAreSet(230, 0) '<--displays: False 230 = 00000000 11100110
MsgBox IntegerBitsAreSet(85, 0, 2, 6, 4) '<--displays: True 85 = 00000000 01010101
MsgBox IntegerBitsAreSet(85, 0, 2, 5, 4) '<--displays: False 85 = 00000000 01010101
MsgBox IntegerBitsAreSet(485, 2, 7, 5) '<--displays: True 485 = 00000001 11100101
MsgBox IntegerBitsAreSet(24585, 14, 13, 3) '<--displays: True 24585 = 01100000 00001001
'
'
'

Source : http://academy.excelhero.com/ | Last Update : Sat, 13 Feb 21

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