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I'm trying to write a function that returns the number of bits set in a 32-bit integer in VBScript, it's just for practising the language. The function I've written so far looks okay, but I'm wondering if there was a simpler solution.

Function countBits(value)
  Dim n : n = 0
  Dim mask : mask = 1
  Dim i
  For i = 0 to 30
    If (value And mask) > 0 Then
      n = n +1
    End If
  Next
  If (value And &h8000) then
    n = n+1
  End If
  countBits = n
End Function

I found that there are no shift operators in VBScript and an overflow (Err.Number = 6) if I iterate for 0 to 31, that's why I add the explicitly check of the MSB after the look.

Any solutions to improve (maybe generalize) this?


edit: removed the useless error handling part, as it's a remains of starting with 31 as an upper bound in the first (and at that time only) loop

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According to information in VBScript Data Types MSDN article, a 32-bit integer is long variant subtype.

However, the value &h8000 (decimal -32768) seems to be a 16-bit integer minimal value; it's hexadecimal &hFFFF8000 if converted to long subtype (check Wscript.Echo Hex(&h8000).

This means that your function miscounts 1 additional bit for all long values above 32767 i.e. above 16-bit integer maximal value &h7FFF.

To check the sign bit of a long value correctly, use &h80000000 as follows:

Function countBits(value)
  Dim mask : mask = 1
  Dim i
  countBits = 0                    ' no need to use any auxiliary variable
  For i = 0 to 30
    If ( value And mask ) > 0 Then
      countBits = countBits + 1
    End If
    mask = mask * 2                ' got lost due to your own edit Dec 11 '16 at 14:22
  Next
  If ( value And &h80000000 ) Then ' check the sign bit
    countBits = countBits + 1
  End If
End Function

Sample tests:

Wscript.Echo countBits(2^16)         '  1
Wscript.Echo countBits(2^30)         '  1
Wscript.Echo countBits(1)            '  1
Wscript.Echo countBits(-1)           ' 32
Wscript.Echo countBits(-2)           ' 31
Wscript.Echo countBits(-2147483648)  '  1
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  • \$\begingroup\$ thanks so much sometimes I seem to be so blind for some important details, the off-by-one syndrome has so many faces this time I was off by one chunk of hexadecimal places. \$\endgroup\$ – Wolf Jul 31 '17 at 14:46

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