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VB字节数组，转成浮点数嘿嘿，还可以吧
fareasy.taobao.com
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Public Function BIN_to_DEC(ByVal Bin As String) As Long
Dim i As Long
For i = 1 To Len(Bin)
BIN_to_DEC = BIN_to_DEC * 2 + Val(Mid(Bin, i, 1))
Next i
End Function
Public Function DEC_to_BIN(Dec As Long) As String
DEC_to_BIN = ""
Do While Dec > 0
DEC_to_BIN = Dec Mod 2 & DEC_to_BIN
Dec = Dec \ 2
Loop
End Function
Private Sub Form_Load()
Dim B As Single
Dim Data(3) As Byte
Dim zhengfu As Integer
Data(0) = &H42 '
Data(1) = &HE4
Data(2) = &H80
Data(3) = &H0
'42E48000 = 114.25
'第一步 , 化为2进制
'0100 0010 1110 0100 1000 0000 0000 0000
'第二步:因为IEEE754使用1个符号,8个阶码,23个尾数,我们分别在上面提取这些内容.
'符号:第一位:0 表示正数
If Data(0) / 128 >= 1 Then
zhengfu = -1
Else
zhengfu = 1
End If
'阶码:2-9位:10000101 为133,实际的幂值为133-127=6
jiema = (Data(0) Mod 128) * 2 + Int(Data(1) / 128) - 127
'尾数:10-32位:11001001000000000000000,实际值为1.11001001 (1+尾数)
'weishu = (Data(1) Mod 128) * 256 * 256 + Data(2) * 256 + Data(3) '小数转数据方式？
Dim weishu As String
weishu = "1" + Format(DEC_to_BIN(Data(1) Mod 128), "0000000") + Format(DEC_to_BIN(Data(2) + 0), "00000000") + Format(DEC_to_BIN(Data(3) + 0), "00000000")
'第三步: 根据公式写出实际数值大小
'为1110010.01,化为十进制为:114.25
result = zhengfu * BIN_to_DEC(weishu) / (2 ^ (15 - jiema))
End Sub
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