ide – 如何在VB6中声明MAX_DOUBLE？

发布时间：2020-12-17 07:17:45 所属栏目：百科来源：网络整理

导读：根据MSDN对VB6的帮助 Floating-point values can be expressed as mmmEeee or mmmDeee,in which mmm is the mantissa and eee is the exponent (a power of 10). The highest positive value of a Single data type is 3.402823E+38,or 3.4 times 10 to the

根据MSDN对VB6的帮助

Floating-point values can be expressed as mmmEeee or mmmDeee,in which mmm is the mantissa and eee is the exponent (a power of 10). The highest positive value of a Single data type is 3.402823E+38,or 3.4 times 10 to the 38th power; the highest positive value of a Double data type is 1.79769313486232D+308,or about 1.8 times 10 to the 308th power. Using D to separate the mantissa and exponent in a numeric literal causes the value to be treated as a Double data type. Likewise,using E in the same fashion treats the value as a Single data type.

现在在VB6 IDE中我试图输入它

const MAX_DOUBLE as Double = 1.79769313486232D+308

但是,一旦我离开该行,IDE就会抛出错误6(溢出)

An overflow results when you try to make an assignment that exceeds the limitations of the target of the assignment. …

那么如何定义MAX_DOUBLE(以及MIN_DOUBLE)？

解决方法

它必须是一个Const吗？通过使用Byte数组中的CopyMemory设置正确的位模式,可以将MAX_DOUBLE的精确值转换为变量.

Private Declare Sub CopyMemory Lib "kernel32" Alias "RtlMoveMemory" (Destination As Any,Source As Any,ByVal Length As Long)

Dim Max As Double
Dim Idx As Long
Dim Bits(0 To 7) As Byte

For Idx = 0 To 5
   Bits(Idx) = 255
Next
Bits(6) = 239 ' = 11101111
Bits(7) = 127

For Idx = 0 To 7
   CopyMemory ByVal VarPtr(Max) + Idx,Bits(Idx),1
Next

Debug.Print Max

编辑：我忘了你也问过MIN_DOUBLE,这更容易.

Dim Min As Double
Dim Bits As Byte

Bits = 1
CopyMemory ByVal VarPtr(Min),Bits,1

Debug.Print Min

（编辑：李大同）

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