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函数小数转分数,限于整除分数

发布: 2007-6-30 18:56 | 作者: admin | 来源: | 查看: 27次 | 进入软件测试论坛讨论

领测软件测试网 @#函数实现小数转分数
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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文章来源于领测软件测试网 https://www.ltesting.net/


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