Question

问题：是否可以在Small和Visual Basic中创建Double Float值？

我一直在尝试在Small / Visual Basic中创建一个双浮点值（就像它们中的两个一样）...
而且我一直都没有运气..我总是以这样的错误终止：

    at System.Decimal..ctor(Double value)
    at System.Decimal.op_Explicit(Double value)
    at Microsoft.SmallBasic.Library.Primitive.op_Implicit(Double value)
    at _SmallBasicProgram._Main()

或者，在Visual Basic中运行：

overflow

<小时/> 那么，有没有办法制作双非整数（十进制）精度浮点数？
代码（我试过）是：

小基础：

var1 = 18446744073709551615
var2 = 1797693134862315800000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

（是的，对于零的数量感到抱歉..它们中有303个。）

并在 Visual Basic中：

Module experiment_doesDoubleFloat_workModule
    Dim var1, var2 As Double
    Sub Main()
        var1 = 18446744073709551615
        var2 = 1797693134862315800000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000
    End Sub
End Module

我搞砸了什么？
我也不知道哪个标签真的适合这个...（除了小基本标签）

Answer 1

这不起作用的原因是因为您有效地为编译器提供了Integer格式的常量，并且编译器正在尝试将您的巨大整数转换为浮点值。这会失败，因为您的＆gt; 300位数值大于适合任何标准数据类型的最大整数。

如果要在代码中分配常量值，则必须采用编译器可以解析的格式，即：

 var1 = 1.8446744073709552E+19
 var2 = 1.7976931348623157E+308

事实上，当你输入：

时，你会注意到

 var1 = 1.8446744073709551615E+19

该值自动转换为：

 var1 = 1.8446744073709552E+19

因为原始值包含的精度高于double格式所能容纳的精度。另外，我没有计算代码示例中的零个数，但如果它是303个，则会产生var2 = 1.797... E+319的值，这对于double来说也太大了。使用292个零时，值变为...E+308，如上所述，这是可表示的最大双精度浮点值。

请注意，只要代码中的值足够小以适合Integer，就可以将整数常量分配给浮点变量。见：

var1 = 9223372036854775807  ' << largest Int64

编译好，但还有一个

var1 = 9223372036854775808

失败。

进一步阅读

What Every Computer Scientist Should Know About Floating-Point Arithmetic

Answer 2

如果用于自动化，请尝试以下操作：

 Public Function REAL_to_DOUBLE(ByVal i3264 As Long) '
        Dim sBit As Integer = 1
        Dim expBits As Integer = 8 'for 32 bits, 11 for 64 bits
        Dim expAux As Integer = 127 'for 32 bts, 1023 for 64 bits
        Dim nBits As Integer = 32 ' for 23 bits, 64 for 64 bits
        Dim dec As Double = 1

        Dim hexstring As String = Hex(i3264)

        If hexstring.Length > 8 Then
            expBits = 11
            nBits = 64
            expAux = 1023
        End If

        Dim bin_ As String
        If nBits = 32 Then
            bin_ = Convert.ToString(Convert.ToInt32(hexstring, 16), 2).PadLeft(nBits, "0"c)
        Else
            bin_ = Convert.ToString(Convert.ToInt64(hexstring, 16), 2).PadLeft(nBits, "0"c)
        End If
        Dim _sinal As Integer = -1
        If (bin_.Substring(0, 1)) = "0" Then _sinal = 1

        Dim _e As String = bin_.Substring(1, expBits).PadLeft(expBits, "0"c)
        Dim a As Integer = Convert.ToInt32(_e, 2)
        Dim exp_ As Integer = a - expAux

        Dim matissa As String = bin_.Substring(expBits + 1, bin_.Length - (expBits + 1))

        Dim length As Integer = Len(matissa)
        Dim ps As Long = 2
        For x As Integer = 0 To length
            Dim temp As Integer = Val(Mid(matissa, x + 1, 1))

            If temp = 1 Then
                dec += temp / ps
            End If
            ps *= 2
        Next

        dec = _sinal * 2 ^ (a - expAux) * dec

        Return dec
    End Function

是否可以制作双浮动值？

2 个答案: