如何将浮点数转换为字节序列，以便可以将其保存在文件中？这种算法必须快速且高度便携。它必须允许相反的操作，反序列化。如果每个值（持久空间）只需要非常微小的多余位，那就太好了。
答案 0 :(得分：12)
假设您正在使用主流编译器，C和C ++中的浮点值遵循IEEE标准，并且当以二进制形式写入文件时可以在任何其他平台中恢复，前提是您使用相同的字节endianess进行写入和读取。所以我的建议是：选择一个选择的结尾，在写作之前或阅读之后，检查该结尾是否与当前平台相同;如果没有，只需交换字节。
答案 1 :(得分：3)
您始终可以按固定字节顺序（小端或大端）转换为IEEE-754格式。对于大多数机器来说，根本不需要任何东西，或者需要简单的字节交换来序列化和反序列化。一台本机不支持IEEE-754的机器需要编写一个转换器，但使用ldexp和frexp（stanard C库函数）和位混洗这样做并不太难。
答案 2 :(得分：3)
这可能会给你一个良好的开端 - 它会将一个浮点值打包到int和long long对中，然后您可以按照常规方式进行序列化。
#define FRAC_MAX 9223372036854775807LL /* 2**63 - 1 */
struct dbl_packed
{
int exp;
long long frac;
};
void pack(double x, struct dbl_packed *r)
{
double xf = fabs(frexp(x, &r->exp)) - 0.5;
if (xf < 0.0)
{
r->frac = 0;
return;
}
r->frac = 1 + (long long)(xf * 2.0 * (FRAC_MAX - 1));
if (x < 0.0)
r->frac = -r->frac;
}
double unpack(const struct dbl_packed *p)
{
double xf, x;
if (p->frac == 0)
return 0.0;
xf = ((double)(llabs(p->frac) - 1) / (FRAC_MAX - 1)) / 2.0;
x = ldexp(xf + 0.5, p->exp);
if (p->frac < 0)
x = -x;
return x;
}
答案 3 :(得分：2)
你是什么意思，“便携式”？
为了便于携带，请记住将数字保持在标准中定义的限制范围内：使用超出这些限制的单个数字，并将所有可移植性排除在外。
double planck_time = 5.39124E-44; /* second */
[...]
10 The values given in the following list shall be replaced by constant
expressions with implementation-defined values [...]
11 The values given in the following list shall be replaced by constant
expressions with implementation-defined values [...]
12 The values given in the following list shall be replaced by constant
expressions with implementation-defined (positive) values [...]
[...]
请注意所有这些条款中的实现定义。
答案 4 :(得分：1)
转换为ascii表示法是最简单的，但是如果你需要处理大量浮点数，那么你当然应该使用二进制文件。但如果您关心可移植性，这可能是一个棘手的问题。浮点数在不同的机器中表示不同。
如果您不想使用固定库，那么您的浮点二进制串行器/解串器只需要在每个位落地及其代表的地方签订“合同”。
这是一个有趣的网站：link。
答案 5 :(得分：0)
sprintf，fprintf？你没有那么便携了。
答案 6 :(得分：0)
您需要什么级别的便携性？如果要在与生成的操作系统相同的计算机上读取文件，则使用二进制文件并保存和恢复位模式应该可以正常工作。否则正如boytheo所说，ASCII是你的朋友。
答案 7 :(得分：0)
此版本每个浮点值只有一个字节，表示字节序。但我认为，它仍然不是很便携。
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
#define LITEND 'L'
#define BIGEND 'B'
typedef short INT16;
typedef int INT32;
typedef double vec1_t;
typedef struct {
FILE *fp;
} WFILE, RFILE;
#define w_byte(c, p) putc((c), (p)->fp)
#define r_byte(p) getc((p)->fp)
static void w_vec1(vec1_t v1_Val, WFILE *p)
{
INT32 i;
char *pc_Val;
pc_Val = (char *)&v1_Val;
w_byte(LITEND, p);
for (i = 0; i<sizeof(vec1_t); i++)
{
w_byte(pc_Val[i], p);
}
}
static vec1_t r_vec1(RFILE *p)
{
INT32 i;
vec1_t v1_Val;
char c_Type,
*pc_Val;
pc_Val = (char *)&v1_Val;
c_Type = r_byte(p);
if (c_Type==LITEND)
{
for (i = 0; i<sizeof(vec1_t); i++)
{
pc_Val[i] = r_byte(p);
}
}
return v1_Val;
}
int main(void)
{
WFILE x_FileW,
*px_FileW = &x_FileW;
RFILE x_FileR,
*px_FileR = &x_FileR;
vec1_t v1_Val;
INT32 l_Val;
char *pc_Val = (char *)&v1_Val;
INT32 i;
px_FileW->fp = fopen("test.bin", "w");
v1_Val = 1234567890.0987654321;
printf("v1_Val before write = %.20f \n", v1_Val);
w_vec1(v1_Val, px_FileW);
fclose(px_FileW->fp);
px_FileR->fp = fopen("test.bin", "r");
v1_Val = r_vec1(px_FileR);
printf("v1_Val after read = %.20f \n", v1_Val);
fclose(px_FileR->fp);
return 0;
}
答案 8 :(得分：0)
我们走了。
便携式IEEE 754序列化/反序列化应该 无论机器的内部浮点如何都能正常工作 表示。
https://github.com/MalcolmMcLean/ieee754
/*
* read a double from a stream in ieee754 format regardless of host
* encoding.
* fp - the stream
* bigendian - set to if big bytes first, clear for little bytes
* first
*
*/
double freadieee754(FILE *fp, int bigendian)
{
unsigned char buff[8];
int i;
double fnorm = 0.0;
unsigned char temp;
int sign;
int exponent;
double bitval;
int maski, mask;
int expbits = 11;
int significandbits = 52;
int shift;
double answer;
/* read the data */
for (i = 0; i < 8; i++)
buff[i] = fgetc(fp);
/* just reverse if not big-endian*/
if (!bigendian)
{
for (i = 0; i < 4; i++)
{
temp = buff[i];
buff[i] = buff[8 - i - 1];
buff[8 - i - 1] = temp;
}
}
sign = buff[0] & 0x80 ? -1 : 1;
/* exponet in raw format*/
exponent = ((buff[0] & 0x7F) << 4) | ((buff[1] & 0xF0) >> 4);
/* read inthe mantissa. Top bit is 0.5, the successive bits half*/
bitval = 0.5;
maski = 1;
mask = 0x08;
for (i = 0; i < significandbits; i++)
{
if (buff[maski] & mask)
fnorm += bitval;
bitval /= 2.0;
mask >>= 1;
if (mask == 0)
{
mask = 0x80;
maski++;
}
}
/* handle zero specially */
if (exponent == 0 && fnorm == 0)
return 0.0;
shift = exponent - ((1 << (expbits - 1)) - 1); /* exponent = shift + bias */
/* nans have exp 1024 and non-zero mantissa */
if (shift == 1024 && fnorm != 0)
return sqrt(-1.0);
/*infinity*/
if (shift == 1024 && fnorm == 0)
{
#ifdef INFINITY
return sign == 1 ? INFINITY : -INFINITY;
#endif
return (sign * 1.0) / 0.0;
}
if (shift > -1023)
{
answer = ldexp(fnorm + 1.0, shift);
return answer * sign;
}
else
{
/* denormalised numbers */
if (fnorm == 0.0)
return 0.0;
shift = -1022;
while (fnorm < 1.0)
{
fnorm *= 2;
shift--;
}
answer = ldexp(fnorm, shift);
return answer * sign;
}
}
/*
* write a double to a stream in ieee754 format regardless of host
* encoding.
* x - number to write
* fp - the stream
* bigendian - set to write big bytes first, elee write litle bytes
* first
* Returns: 0 or EOF on error
* Notes: different NaN types and negative zero not preserved.
* if the number is too big to represent it will become infinity
* if it is too small to represent it will become zero.
*/
int fwriteieee754(double x, FILE *fp, int bigendian)
{
int shift;
unsigned long sign, exp, hibits, hilong, lowlong;
double fnorm, significand;
int expbits = 11;
int significandbits = 52;
/* zero (can't handle signed zero) */
if (x == 0)
{
hilong = 0;
lowlong = 0;
goto writedata;
}
/* infinity */
if (x > DBL_MAX)
{
hilong = 1024 + ((1 << (expbits - 1)) - 1);
hilong <<= (31 - expbits);
lowlong = 0;
goto writedata;
}
/* -infinity */
if (x < -DBL_MAX)
{
hilong = 1024 + ((1 << (expbits - 1)) - 1);
hilong <<= (31 - expbits);
hilong |= (1 << 31);
lowlong = 0;
goto writedata;
}
/* NaN - dodgy because many compilers optimise out this test, but
*there is no portable isnan() */
if (x != x)
{
hilong = 1024 + ((1 << (expbits - 1)) - 1);
hilong <<= (31 - expbits);
lowlong = 1234;
goto writedata;
}
/* get the sign */
if (x < 0) { sign = 1; fnorm = -x; }
else { sign = 0; fnorm = x; }
/* get the normalized form of f and track the exponent */
shift = 0;
while (fnorm >= 2.0) { fnorm /= 2.0; shift++; }
while (fnorm < 1.0) { fnorm *= 2.0; shift--; }
/* check for denormalized numbers */
if (shift < -1022)
{
while (shift < -1022) { fnorm /= 2.0; shift++; }
shift = -1023;
}
/* out of range. Set to infinity */
else if (shift > 1023)
{
hilong = 1024 + ((1 << (expbits - 1)) - 1);
hilong <<= (31 - expbits);
hilong |= (sign << 31);
lowlong = 0;
goto writedata;
}
else
fnorm = fnorm - 1.0; /* take the significant bit off mantissa */
/* calculate the integer form of the significand */
/* hold it in a double for now */
significand = fnorm * ((1LL << significandbits) + 0.5f);
/* get the biased exponent */
exp = shift + ((1 << (expbits - 1)) - 1); /* shift + bias */
/* put the data into two longs (for convenience) */
hibits = (long)(significand / 4294967296);
hilong = (sign << 31) | (exp << (31 - expbits)) | hibits;
x = significand - hibits * 4294967296;
lowlong = (unsigned long)(significand - hibits * 4294967296);
writedata:
/* write the bytes out to the stream */
if (bigendian)
{
fputc((hilong >> 24) & 0xFF, fp);
fputc((hilong >> 16) & 0xFF, fp);
fputc((hilong >> 8) & 0xFF, fp);
fputc(hilong & 0xFF, fp);
fputc((lowlong >> 24) & 0xFF, fp);
fputc((lowlong >> 16) & 0xFF, fp);
fputc((lowlong >> 8) & 0xFF, fp);
fputc(lowlong & 0xFF, fp);
}
else
{
fputc(lowlong & 0xFF, fp);
fputc((lowlong >> 8) & 0xFF, fp);
fputc((lowlong >> 16) & 0xFF, fp);
fputc((lowlong >> 24) & 0xFF, fp);
fputc(hilong & 0xFF, fp);
fputc((hilong >> 8) & 0xFF, fp);
fputc((hilong >> 16) & 0xFF, fp);
fputc((hilong >> 24) & 0xFF, fp);
}
return ferror(fp);
}
答案 9 :(得分：-1)
fwrite（），fread（）？你可能会想要二进制文件，除非你想牺牲你在程序中执行的精度然后fwrite（）fread（），否则你不能将字节打包得更紧。漂浮一个;双b; α=（浮点）B;的fwrite（＆安培; A，1，的sizeof（a）中，FP）;
如果您携带不同的浮点格式，它们可能无法以直接的二进制方式进行转换，因此您可能必须分开这些位并执行数学计算，此功能加上此功能，等等.IEEE 754是一个可怕的标准使用但广泛，所以它会尽量减少努力。