[英]How do I convert a double to a proprietary floating point format in C#?
我有一种20位专有浮点格式,定义为:
Bits: 19 18 17 16 15 14 13 12 11 10 09 08 07 06 05 04 03 02 01 00
Use: s e e e e e m m m m m m m m m m m m m m
Where
s = sign, (1=negative, 0=positive)
e = exponent, (10^0 to 10^31)
m = mantissa, (2^0 to 2^14-1, 0 to 16383)
该位分配给出了从-16383 X 10 ^ 31到+16383 X 10 ^ 31的值范围。
现在,如果我在C#中有一个双精度值(最好是十进制),如何将这个数字转换为这种专有的浮点格式? 有一个十进制构造函数Decimal(int lo, int mid, int hi, bool isNegative, byte scale)
-可以帮助我将这种格式转换为十进制,但不能Decimal(int lo, int mid, int hi, bool isNegative, byte scale)
。 交给你...
这是进行转换的一种方法(可能不是最佳方法):
public static class Converter
{
private const int cSignBit = 0x80000; // 1 bit
private const int cExponentBits = 0x7C000; // 5 bits
private const int cMaxExponent = 0x1F; // 5 bits
private const int cMantissaBits = 0x03FFF; // 14 bits
private const double cExponentBase = 10;
private const int cScale = 100000;
public static int DoubleToAmount(double aValue)
{
// Get the sign
bool lNegative = (aValue < 0);
// Get the absolute value with scaling
double lValue = Math.Abs(aValue) * cScale;
// Now keep dividing by 10 to get the exponent value
int lExponent = 0;
while (Math.Ceiling(lValue) > cMantissaBits)
{
lExponent++;
if (lExponent > cMaxExponent)
throw new ArgumentOutOfRangeException("Cannot convert amount", (Exception)null);
lValue = lValue / (double)cExponentBase;
}
// The result that is left rounded up is the mantissa
int lMantissa = (int)Math.Ceiling(lValue);
// Now we pack it into the specified format
int lAmount = (lNegative ? cSignBit : 0) + lExponent * (cMantissaBits + 1) + lMantissa;
// And return the result
return lAmount;
}
}
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