我该怎么做定点分割?

时间:2012-10-13 19:29:02

标签: c# fixed-point

我使用longs得到一个8字节的定点数,其恒定分母为(1 <&lt;&lt;&lt;&lt;&lt;&lt;&lt; 24)。如何分割两个Fixed8值?由于我使用C#,我不能简单地转换为更大的整数。或者我在语言上想得太多了?

感谢。

public struct Fixed8
{
    private long _numerator;
    public const long DENOMINATOR = 1 << 24;
}

4 个答案:

答案 0 :(得分:1)

您可以使用BigInteger Structure执行计算:

public static Fixed8 operator /(Fixed8 a, Fixed8 b)
{
    Fixed8 result;
    result._numerator = (long)( new BigInteger(a._numerator) *
                                new BigInteger(DENOMINATOR)  /
                                new BigInteger(b._numerator) );
    return result;
}

完整代码:

public const long DENOMINATOR = 1 << 24;

private long _numerator;

public Fixed8(double value)
{
    _numerator = (long)(value * DENOMINATOR);
}

public static Fixed8 operator /(Fixed8 a, Fixed8 b)
{
    Fixed8 result;
    result._numerator = (long)( new BigInteger(a._numerator) * 
                                new BigInteger(DENOMINATOR)  / 
                                new BigInteger(b._numerator) );
    return result;
}

public static explicit operator double(Fixed8 a)
{
    return (double)a._numerator / (double)DENOMINATOR;
}

public override string ToString()
{
    return ((double)this).ToString();
}

示例:

var a = new Fixed8(7);
var b = new Fixed8(1.7);

Console.WriteLine(a);
Console.WriteLine(b);
Console.WriteLine(a / b);

输出:

7
1.69999998807907
4.11764705181122

答案 1 :(得分:1)

这是一种将所有计算保持在long s的方法。但是,它可能不会更快;我没有测量过。

public struct Fixed8
{
    public Fixed8(double value)
    {
        _numerator = (long)(value * DENOMINATOR);
    }

    private long _numerator;
    public const long DENOMINATOR = 1 << 24;

    public static Fixed8 operator /(Fixed8 a, Fixed8 b)
    {
        long remainder;
        long quotient = Math.DivRem(a._numerator, b._numerator, out remainder) * DENOMINATOR;

        long morePrecision = remainder * DENOMINATOR / b._numerator;

        return new Fixed8 { _numerator = quotient + morePrecision };
    }
}

答案 2 :(得分:0)

public long Division()
{
    long result = _numerator / DENOMINATOR;
    return result;
}

答案 3 :(得分:0)

使用System.Numerics.BigInteger很好。但在这种特定情况下,System.Decimal实际上具有足够的精度。所以这是我的建议:

 public static Fixed8 operator /(Fixed8 a, Fixed8 b)
 {
   decimal resultNumerator = (decimal)a._numerator * DENOMINATOR / b._numerator;
   return new Fixed8 { _numerator = Convert.ToInt64(resultNumerator) };
 }