这是一个解释解码的 cmets 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 C 库函数。 ldexp(x, e) 返回 x 乘以 2 的 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printf 转换规范，用于格式化uint32_t。