我发现了以下问题Convert Delphi Real48 to 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 []）。
由于
答案 0 :(得分：5)
我遇到了这个寻找相同代码的线程。以下是我最后写的内容：
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。
备注： 我认为这段代码不适用于大端机器。它不检查NAN或INF。
以下是将real48转换为double的代码。它来自Free Pascal编译器：
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);
}
答案 1 :(得分：1)
如果您熟悉C（因为您使用C＃编写应该没问题），请查看此函数。把它移到C＃中应该不会太困难。
这是相当丑陋的，但我想的是必要的。
参考：http://forums.ni.com/ni/board/message?board.id=60&message.id=3553
enum prconverr double_to_real (double d, real *r)
/* converts C double to Pascal real, returns error code */
{
union doublearray da;
unsigned x;
da.d = d;
/* check for 0.0 */
if ((da.a[0] == 0x0000) &&
(da.a[1] == 0x0000) &&
(da.a[2] == 0x0000) &&
/* ignore sign bit */
((da.a[3] & 0x7FFF) == 0x0000)) {
/* exponent and significand are both 0, so value is 0.0 */
(*r)[2] = (*r)[1] = (*r)[0] = 0x0000;
/* sign bit is ignored ( -0.0 -> 0.0 ) */
return prOK;
}
/* test for maximum exponent value */
if ((da.a[3] & 0x7FF0) == 0x7FF0) {
/* value is either Inf or NaN */
if ((da.a[0] == 0x0000) &&
(da.a[1] == 0x0000) &&
(da.a[2] == 0x0000) &&
((da.a[3] & 0x000F) == 0x0000)) {
/* significand is 0, so value is Inf */
/* value becomes signed maximum real, */
/* and error code prInf is returned */
(*r)[1] = (*r)[0] = 0xFFFF;
(*r)[2] = 0x7FFF |
(da.a[3] & 0x8000); /* retain sign bit */
return prInf;
} else {
/* significand is not 0, so value is NaN */
/* value becomes 0.0, and prNaN code is returned */
/* sign bit is ignored (no negative NaN) */
(*r)[2] = (*r)[1] = (*r)[0] = 0x0000;
/* sign bit is ignored ( -NaN -> +NaN ) */
return prNaN;
}
}
/* round significand if necessary */
if ((da.a[0] & 0x1000) == 0x1000) {
/* significand's 40th bit set, so round significand up */
if ((da.a[0] & 0xE000) != 0xE000)
/* room to increment 3 most significant bits */
da.a[0] += 0x2000;
else {
/* carry bit to next element */
da.a[0] = 0x0000;
/* carry from 0th to 1st element */
if (da.a[1] != 0xFFFF)
da.a[1]++;
else {
da.a[1] = 0x0000;
/* carry from 1st to 2nd element */
if (da.a[2] != 0xFFFF)
da.a[2]++;
else {
da.a[2] = 0x0000;
/* carry from 2nd to 3rd element */
/* significand may overflow into exponent */
/* exponent not full, so won't overflow */
da.a[3]++;
}
}
}
}
/* get exponent for underflow/overflow tests */
x = (da.a[3] & 0x7FF0) >> 4;
/* test for underflow */
if (x < 895) {
/* value is below real range */
(*r)[2] = (*r)[1] = (*r)[0] = 0x0000;
if ((da.a[3] & 0x8000) == 0x8000)
/* sign bit was set, so value was negative */
return prNegUnderflow;
else
/* sign bit was not set */
return prPosUnderflow;
}
/* test for overflow */
if (x > 1149) {
/* value is above real range */
(*r)[1] = (*r)[0] = 0xFFFF;
(*r)[2] = 0x7FFF | (da.a[3] & 0x8000); /* retain sign bit */
return prOverflow;
}
/* value is within real range */
(*r)[0] = (x - 894) | /* re-bias exponent */
((da.a[0] & 0xE000) >> 5) | /* begin significand */
(da.a[1] << 11);
(*r)[1] = (da.a[1] >> 5) |
(da.a[2] << 11);
(*r)[2] = (da.a[2] >> 5) |
((da.a[3] & 0x000F) << 11) |
(da.a[3] & 0x8000); /* copy sign bit */
return prOK;
}
答案 2 :(得分：1)
如果可能的话，最简单的方法是将其转换为字符串，传递，然后将其转换回Real48
答案 3 :(得分：0)
double Double_Real48(double d)
{
unsigned long long r48 = 0, tmp;
tmp = *(long long *)&d;//m
tmp/=0x20;
tmp&=0x7FFFFFFFFF00;
r48+=tmp;
tmp = *(long long *)&d;//e
tmp/=0x10000000000000;
tmp-=894;
tmp&=0xFF;
if (tmp == 0) return 0.0;
r48+=tmp;
tmp = *(long long *)&d;//s
tmp/=0x10000;
tmp&=0x800000000000;
r48+=tmp;
return *(double *)&r48;
}
double Real48_Double(double r48)
{
unsigned long long d = 0, tmp;
tmp= *(long long *)&r48;//m
tmp&=0x7FFFFFFFFF00;
tmp*=0x20;
d+=tmp;
tmp= *(long long *)&r48;//e
tmp&=0xFF;
if (tmp == 0) return 0.0;
tmp+=894;
tmp*=0x10000000000000;
d+=tmp;
tmp= *(long long *)&r48;//s
tmp&=0x800000000000;
tmp*=0x10000;
d+=tmp;
return *(double *)&d;
}
答案 4 :(得分：0)
在C / C ++中
typedef struct {
unsigned char exponent; // 8 bites;
unsigned long mantisaLo; // 32 of 39 bites
unsigned char mantisaHi : 7, sign : 1; // 7 of 39 bites
} T_Real48;
typedef struct {
unsigned long mantisaLo; // 32 of 52 bites
unsigned long mantisaHi:20, exponent: 11, sign : 1; // 20 of 52 bites
} T_Double64;
double doubleToReal48(double val)
{
T_Real48 real48;
T_Double64 *double64 = (T_Double64*) &val;
real48.mantisaHi = double64->mantisaHi >> 13;
real48.mantisaLo =(double64->mantisaLo >> 13) + ((double64->mantisaHi & 0x1FFF) << 19);
real48.exponent = double64->exponent - 894;
real48.sign = double64->sign;
if (real48.exponent == 0) {
real48.mantisaHi = 0;
real48.mantisaLo = 0;
}
return *(double *)&real48;
}
double real48ToDouble(double val)
{
T_Real48 *real48 = (T_Real48*) &val;
T_Double64 double64;
double64.mantisaHi = (real48->mantisaHi << 13) + (real48->mantisaLo >> 19);
double64.mantisaLo = real48->mantisaLo << 13;
double64.exponent = real48->exponent + 894;
double64.sign = real48->sign;
return *(double *)&double64;
}