将C#double转换为Delphi Real48

Question

我发现以下问题将Delphi Real48转换为C#double,但我想转向其他方式,C#转向Delphi.

有谁知道如何做到这一点？我试过逆向工程代码,但没有太多运气.

更新:

我正在使用C#代码,它将采用double并将其转换为Real48(大小为6的byte []).

谢谢

Answer 1

我遇到了这个寻找相同代码的线程.这是我最后写的:

public static byte [] Double2Real48(double d)
{
    byte [] r48 = new byte[6];
    byte [] da = BitConverter.GetBytes(d);

    for (int i = 0; i < r48.Length; i++)
        r48[i] = 0;

    //Copy the negative flag
    r48[5] |= (byte)(da[7] & 0x80);

    //Get the expoent
    byte b1 = (byte)(da[7] & 0x7f);
    ushort n = (ushort)(b1 << 4);
    byte b2 = (byte)(da[6] & 0xf0);
    b2 >>= 4;
    n |= b2;

    if (n == 0)
        return r48;

    byte ex = (byte)(n - 1023);
    r48[0] = (byte)(ex + 129);

    //Copy the Mantissa
    r48[5] |= (byte)((da[6] & 0x0f) << 3);//Get the last four bits
    r48[5] |= (byte)((da[5] & 0xe0) >> 5);//Get the first three bits

    r48[4]  = (byte)((da[5] & 0x1f) << 3);//Get the last 5 bits
    r48[4] |= (byte)((da[4] & 0xe0) >> 5);//Get the first three bits

    r48[3]  = (byte)((da[4] & 0x1f) << 3);//Get the last 5 bits
    r48[3] |= (byte)((da[3] & 0xe0) >> 5);//Get the first three bits

    r48[2]  = (byte)((da[3] & 0x1f) << 3);//Get the last 5 bits
    r48[2] |= (byte)((da[2] & 0xe0) >> 5);//Get the first three bits

    r48[1]  = (byte)((da[2] & 0x1f) << 3);//Get the last 5 bits
    r48[1] |= (byte)((da[1] & 0xe0) >> 5);//Get the first three bits

    return r48;

}

Real48类似于IEEE 754,因为尾数将是相同的.位移是必要的,以使尾数位于正确的位置.

Real48指数偏差为129,双精度偏差为1023.

负标志存储在最后一个字节的第一位.

注意: 我认为此代码不适用于大端机器.它不检查NAN或INF.

这是将real48转换为double的代码.它是从Free Pascal编译器移植的:

static double real2double(byte [] r)
{
    byte [] res = new byte[8];
    int exponent;

    //Return zero if the exponent is zero        
    if (r[0] == 0)
        return (double)0;

    //Copy Mantissa
    res[0] = 0;
    res[1] = (byte)(r[1] << 5);
    res[2] = (byte)((r[1] >> 3) | (r[2] << 5));
    res[3] = (byte)((r[2] >> 3) | (r[3] << 5));
    res[4] = (byte)((r[3] >> 3) | (r[4] << 5));
    res[5] = (byte)((r[4] >> 3) | ((r[5] & 0x7f) << 5));
    res[6] = (byte)((r[5] & 0x7f) >> 3);

    //Copy exponent
    //correct exponent
    exponent = (r[0] + (1023-129));
    res[6] = (byte)(res[6] | ((exponent & 0xf) << 4));
    res[7] = (byte)(exponent >> 4);

    //Set Sign
    res[7] = (byte)(res[7] | (r[5] & 0x80));
    return BitConverter.ToDouble(res, 0);  
}