似乎VB6不太容易将+无限大、-无限大和NaN存储到double变量中。如果可以这样做,我可以在复数的上下文中使用这些值进行比较。该如何实现?
实际上,有一种更简单的方法可以获得无穷大、负无穷大和非数值:
public lfNaN as Double ' or As Single
public lfPosInf as Double
public lfNegInf as Double
on error resume next ' to ignore Run-time error '6': Overflow and '11': Division by zero
lfNaN = 0 / 0 ' -1.#IND
lfPosInf = 1 / 0 ' 1.#INF
lfNegInf = -1 / 0 ' -1.#INF
on error goto 0 ' optional to reset the error handler
有几件不同的事情。就像你从Pax的例子中看到的那样,你只需要查找IEEE 754标准,然后将字节放入正确的位置即可。我唯一要提醒你的是,微软已经弃用RtlMoveMemory,因为它可能会创建溢出类型的安全问题。作为替代方案,你可以使用用户定义类型和LSet在“纯”VB中进行小心的强制转换实现这一点。(还请注意,有两种类型的NaN。)
Option Explicit
Public Enum abIEEE754SpecialValues
abInfinityPos
abInfinityNeg
abNaNQuiet
abNaNSignalling
abDoubleMax
abDoubleMin
End Enum
Private Type TypedDouble
value As Double
End Type
Private Type ByteDouble
value(7) As Byte
End Type
Public Sub Example()
MsgBox GetIEEE754SpecialValue(abDoubleMax)
End Sub
Public Function GetIEEE754SpecialValue(ByVal value As abIEEE754SpecialValues) As Double
Dim dblRtnVal As Double
Select Case value
Case abIEEE754SpecialValues.abInfinityPos
dblRtnVal = BuildDouble(byt6:=240, byt7:=127)
Case abIEEE754SpecialValues.abInfinityNeg
dblRtnVal = BuildDouble(byt6:=240, byt7:=255)
Case abIEEE754SpecialValues.abNaNQuiet
dblRtnVal = BuildDouble(byt6:=255, byt7:=255)
Case abIEEE754SpecialValues.abNaNSignalling
dblRtnVal = BuildDouble(byt6:=248, byt7:=255)
Case abIEEE754SpecialValues.abDoubleMax
dblRtnVal = BuildDouble(255, 255, 255, 255, 255, 255, 239, 127)
Case abIEEE754SpecialValues.abDoubleMin
dblRtnVal = BuildDouble(255, 255, 255, 255, 255, 255, 239, 255)
End Select
GetIEEE754SpecialValue = dblRtnVal
End Function
Public Function BuildDouble( _
Optional byt0 As Byte = 0, _
Optional byt1 As Byte = 0, _
Optional byt2 As Byte = 0, _
Optional byt3 As Byte = 0, _
Optional byt4 As Byte = 0, _
Optional byt5 As Byte = 0, _
Optional byt6 As Byte = 0, _
Optional byt7 As Byte = 0 _
) As Double
Dim bdTmp As ByteDouble, tdRtnVal As TypedDouble
bdTmp.value(0) = byt0
bdTmp.value(1) = byt1
bdTmp.value(2) = byt2
bdTmp.value(3) = byt3
bdTmp.value(4) = byt4
bdTmp.value(5) = byt5
bdTmp.value(6) = byt6
bdTmp.value(7) = byt7
LSet tdRtnVal = bdTmp
BuildDouble = tdRtnVal.value
End Function
最后一点需要注意的是,你也可以通过这种方式得到NaN:
Public Function GetNaN() As Double
On Error Resume Next
GetNaN = 0 / 0
End Function
这个页面展示了一种略微复杂的方法来实现它。我已经简化了它以匹配您的问题,但还没有进行彻底测试。如果有任何问题,请告诉我。我注意到该网站上的一个问题是,他们用于静默NaN的代码是错误的,应该使用1位来启动尾数 - 他们似乎将其与信号NaN混淆了。
Public NegInfinity As Double
Public PosInfinity As Double
Public QuietNAN As Double
Private Declare Sub CopyMemoryWrite Lib "kernel32" Alias "RtlMoveMemory" ( _
ByVal Destination As Long, source As Any, ByVal Length As Long)
' IEEE754 doubles: '
' seeeeeee eeeemmmm mmmmmmmm mmmmmmmm mmmmmmmm mmmmmmmm mmmmmmmm mmmmmmmm '
' s = sign '
' e = exponent '
' m = mantissa '
' Quiet NaN: s = x, e = all 1s, m = 1xxx... '
' +Inf : s = 0, e = all 1s, m = all 0s. '
' -Inf : s = 1, e = all 1s, m = all 0s. '
Public Sub Init()
Dim ptrToDouble As Long
Dim byteArray(7) As Byte
Dim i As Integer
byteArray(7) = &H7F
For i = 0 To 6
byteArray(i) = &HFF
Next
ptrToDouble = VarPtr(QuietNAN)
CopyMemoryWrite ptrToDouble, byteArray(0), 8
byteArray(7) = &H7F
byteArray(6) = &HF0
For i = 0 To 5
byteArray(i) = 0
Next
ptrToDouble = VarPtr(PosInfinity)
CopyMemoryWrite ptrToDouble, byteArray(0), 8
byteArray(7) = &HFF
byteArray(6) = &HF0
For i = 0 To 5
byteArray(i) = 0
Next
ptrToDouble = VarPtr(NegInfinity)
CopyMemoryWrite ptrToDouble, byteArray(0), 8
End Sub
它基本上使用内核级内存复制将位模式从字节数组传输到双精度浮点数。
然而,您需要记住,有多个比特值可以表示QNaN,具体来说,符号位可以是0或1,除第一个位以外的所有尾数位也可以是零或1。这可能会使您的比较策略变得复杂,除非您可以发现VB6只使用其中一种位模式-但这不会影响这些值的初始化,假设VB6正确实现了IEE754双精度浮点数。