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问题描述
我正在尝试将double转换为pascal real,但是当我将0.23转换为real时,我却得到0.23999999 real,我如何将所有9999截断为0000。
i am trying to convert double to pascal real but when i convert 0.23 to real i got 0.23999999 real how can i truncate all 9999 to 0000.
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;
}
推荐答案
我的转换程序。请注意,我在这里未对处理IEEE无限性或NaN值做任何特殊说明。
Here are my conversion routines. Note that I haven't put anything special here for treating IEEE infinities or NaN values.
static byte[] DoubleToReal48(double d)
{
byte[] r = new byte[6];
long bits = BitConverter.DoubleToInt64Bits(d);
bool negative = ((bits >> 63) & 1) != 0;
long exponent = ((bits >> 52) & 0x7FF) - 1023;
long mantissa = bits & 0xFFFFFFFFFFFFFL;
long raw = (negative ? 1 : 0);
raw = (raw << 39) | (mantissa >> 13);
raw = (raw << 8) | ((exponent + 129) & 0xFF);
for (int k = 0; k < 6; k++)
{
r[k] = (byte)(raw & 0xFF);
raw >>= 8;
}
return r;
}
static double Real48ToDouble(byte[] r)
{
long raw = 0;
for (int k = 5; k >= 0; k--)
{
raw = (raw << 8) | r[k];
}
long mantissa = (raw << 5) & 0xFFFFFFFFFD000L;
long exponent = (((raw & 0xFF) - 129 + 1023) & 0x7FF) << 52;
long sign = (((raw & ~0x7FFFFFFFFFFFL) != 0) ? 1 : 0) << 63;
return BitConverter.Int64BitsToDouble(sign | exponent | mantissa);
}
往返转换会损失一些精度,但结果基本上是正确的。 [Real48ToDouble(DoubleToReal48(0.23))返回0.229999999999563]
There is some loss of precision with roundtrip conversions, but the results are basically correct. [Real48ToDouble(DoubleToReal48(0.23)) returns 0.229999999999563]
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