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如何编写一个小数转换分数的函数?
摘要:PublicFunctionXtoF(strAsCurrency,OptionalfenmAsInteger=32)AsString'只限于...

Public Function XtoF(str As Currency, Optional fenm As Integer = 32) As String

' 只限于整除分数.

Dim Cfm As Currency

Dim cfmmod As Integer

On Error GoTo Erroreof

Cfm = 1 / fenm

XtoF = ""

If str = 0 Then XtoF = "": Exit Function

Dim point As Integer

Dim dInt As String

Dim dPoint As Currency

Dim fint, fint1, fint2 As Integer

If str <> 0 Then

If str > 1 Then

point = InStr(1, str, ".", 1)

If point = 0 Then

XtoF = str:

Exit Function

Else

dInt = Mid(str, 1, point - 1)

dPoint = CCur("0." & Mid(str, point + 1))

fint = InStr(1, XtoF(dPoint), "/", 1)

fint1 = CInt(Mid(XtoF(dPoint), 1, fint - 1))

fint2 = CInt(Mid(XtoF(dPoint), fint + 1))

XtoF = CStr(dInt * fint2 + fint1) & "/" & CStr(fint2)

End If

Else

If fenm Mod CInt(str / Cfm) = 0 Then

XtoF = "1/" + CStr(fenm / CInt(str / Cfm))

Else

cfmmod = Maxgys(fenm, CInt(str / Cfm))

XtoF = CStr(CInt(str / Cfm / cfmmod)) + "/" + CStr(CInt(fenm / cfmmod))

End If

End If

Else

XtoF = "0"

End If

Exit Function

Erroreof:

XtoF = ""

End Function

Function Maxgys(num1 As Integer, num2 As Integer) As Integer

Dim minnum, i As Integer

minnum = num1

If num1 > num2 Then minnum = num2

For i = 1 To minnum

If ((num1 Mod i) = 0) And ((num2 Mod i) = 0) Then Maxgys = i

Next i

End Function

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