我得到了尾数，指数和符号，我必须将其转换为相应的浮点数。我将22位用于尾数，将9位用于指数，将1位用于符号。
我从概念上知道如何将它们转换为浮点数，首先将指数调整回其位置，然后将结果数字转换回浮点数，但是我在用C实现时遇到了麻烦。我看到了this thread，但是我无法理解代码，而且不确定答案是否正确。谁能指出我正确的方向？我需要在C中进行编码
编辑：通过将尾数转换为二进制，然后调整二进制的小数点，然后将小数点的二进制转换回实际的浮点数，我已经取得了一些进展。我基于这两个GeekforGeek页面（one，two）建立了转换函数，但是似乎所有这些二进制转换都是漫长而艰难的过程。上面的链接显然是通过使用>>运算符完成的，只需很少的步骤，但是我不完全了解这是如何导致浮动的。
这里是带有解释解码的注释的程序：
#include <inttypes.h>
#include <math.h>
#include <stdint.h>
#include <stdio.h>
#include <stdlib.h>
// Define constants describing the floating-point encoding.
enum
{
SignificandBits = 22, // Number of bits in signficand field.
ExponentBits = 9, // Number of bits in exponent field.
ExponentMaximum = (1 << ExponentBits) - 1,
ExponentBias = (1 << ExponentBits-1) - 1,
};
/* Given the contents of the sign, exponent, and significand fields that
encode a floating-point number following IEEE-754 patterns for binary
floating-point, return the encoded number.
"double" is used for the return type as not all values represented by the
sample format (9 exponent bits, 22 significand bits) will fit in a "float"
when it is the commonly used IEEE-754 binary32 format.
*/
double DecodeCustomFloat(
unsigned SignField, uint32_t ExponentField, uint32_t SignificandField)
{
/* We are given a significand field as an integer, but it is used as the
value of a binary numeral consisting of “.” followed by the significand
bits. That value equals the integer divided by 2 to the power of the
number of significand bits. Define a constant with that value to be
used for converting the significand field to represented value.
*/
static const double SignificandRatio = (uint32_t) 1 << SignificandBits;
/* Decode the sign field:
If the sign bit is 0, the sign is +, for which we use +1.
If the sign bit is 1, the sign is -, for which we use -1.
*/
double Sign = SignField ? -1. : +1.;
// Dispatch to handle the different categories of exponent field.
switch (ExponentField)
{
/* When the exponent field is all ones, the value represented is a
NaN or infinity:
If the significand field is zero, it is an infinity.
Otherwise, it is a NaN. In either case, the sign should be
preserved.
Note this is a simple demonstration implementation that does not
preserve the bits in the significand field of a NaN -- we just
return the generic NAN without attempting to set its significand
bits.
*/
case ExponentMaximum:
{
return Sign * (SignificandField ? NAN : INFINITY);
}
/* When the exponent field is not all zeros or all ones, the value
represented is a normal number:
The exponent represented is ExponentField - ExponentBias, and
the significand represented is the value given by the binary
numeral “1.” followed by the significand bits.
*/
default:
{
int Exponent = ExponentField - ExponentBias;
double Significand = 1 + SignificandField / SignificandRatio;
return Sign * ldexp(Significand, Exponent);
}
/* When the exponent field is zero, the value represented is subnormal:
The exponent represented is 1 - ExponentBias, and the
significand represented is the value given by the binary
numeral “0.” followed by the significand bits.
*/
case 0:
{
int Exponent = 1 - ExponentBias;
double Significand = 0 + SignificandField / SignificandRatio;
return Sign * ldexp(Significand, Exponent);
}
}
}
/* Test that a given set of fields decodes to the expected value and
print the fields and the decoded value.
*/
static void Demonstrate(
unsigned SignField, uint32_t SignificandField, uint32_t ExponentField,
double Expected)
{
double Observed
= DecodeCustomFloat(SignField, SignificandField, ExponentField);
if (! (Observed == Expected) && ! (isnan(Observed) && isnan(Expected)))
{
fprintf(stderr,
"Error, expected (%u, %" PRIu32 ", %" PRIu32 ") to represent "
"%g (hexadecimal %a) but got %g (hexadecimal %a).\n",
SignField, SignificandField, ExponentField,
Expected, Expected,
Observed, Observed);
exit(EXIT_FAILURE);
}
printf(
"(%u, %" PRIu32 ", %" PRIu32 ") represents %g (hexadecimal %a).\n",
SignField, SignificandField, ExponentField, Observed, Observed);
}
int main(void)
{
Demonstrate(0, 0, 0, +0.);
Demonstrate(1, 0, 0, -0.);
Demonstrate(0, 255, 0, +1.);
Demonstrate(1, 255, 0, -1.);
Demonstrate(0, 511, 0, +INFINITY);
Demonstrate(1, 511, 0, -INFINITY);
Demonstrate(0, 511, 1, +NAN);
Demonstrate(1, 511, 1, -NAN);
Demonstrate(0, 0, 1, +0x1p-276);
Demonstrate(1, 0, 1, -0x1p-276);
Demonstrate(0, 255, 1, +1. + 0x1p-22);
Demonstrate(1, 255, 1, -1. - 0x1p-22);
Demonstrate(0, 1, 0, +0x1p-254);
Demonstrate(1, 1, 0, -0x1p-254);
Demonstrate(0, 510, 0x3fffff, +0x1p256 - 0x1p233);
Demonstrate(1, 510, 0x3fffff, -0x1p256 + 0x1p233);
}
一些注意事项：
ldexp是标准的C库函数。 ldexp(x, e)返回x乘以2等于e的幂。uint32_t是无符号的32位整数类型。它在stdint.h中定义。"%" PRIu32提供用于格式化printf的uint32_t转换规范。链接的问题不是C ++，而是C。要在C位保留的数据类型之间进行转换，要使用的工具是联合。有点像
union float_or_int {
uint32_t i;
float f;
}
float to_float(int mantissa, int exponent, int sign)
{
union float_or_int result;
result.i = (sign << 31) | (exponent << 22) | mantissa;
return result.f;
}
很抱歉打错，自从我用C编码以来已经有一段时间了