如何将 IEEE 754 单精度二进制浮点数转换为十进制数? [英] How to convert an IEEE 754 single-precision binary floating-point to decimal?

问题描述

我正在开发一个需要将 32 位数字转换为十进制数的程序.

I am working on a program that needs to convert a 32-bit number into a decimal number.

我从输入中得到的数字是一个 32 位数字，表示为浮点数.第一位是符号，接下来的 8 位是指数，其他 23 位是尾数.我正在用 C 语言编写程序.在输入中，我将该数字作为 char[] 数组，然后我在其中创建一个新的 int[] 数组存储符号、指数和尾数.但是，当我尝试将尾数存储为某种数据类型时，我遇到了尾数问题，因为我需要将尾数用作数字，而不是数组: formula=sign*(1+0.mantissa)*2^(exponent-127).

The number that I get from input is a 32 bit number represented as floating point. The first bit is the sign, the next 8 bits are the exponent, and the other 23 bits are mantissa. I am working the program in C. In input, I get that number as a char[] array, and after that I am making a new int[] array where I store the sign , the exponent and the mantissa. But, I have problem with the mantissa when I am trying to store it in some datatype, because I need to use the mantissa as a number, not as an array: formula=sign*(1+0.mantissa)*2^(exponent-127).

这是我用来存储尾数的代码，但程序仍然给我错误的结果:

Here is the code I use to store the mantissa, but still the program gets me wrong results:

double oMantissa=0;
int counter=0;
for(counter=0;counter<23;counter++)
{
    if(mantissa[counter]==1)
    {
        oMantissa+=mantissa[counter]*pow(10,-counter);
    }
}

mantissa[] 是一个 int 数组，我已经从 char 数组转换了尾数.当我从 formula 得到值时，它必须是一个二进制数，我必须将它转换为十进制，所以我会得到这个数字的值.你能帮我存储尾数的 23 位吗?而且，我不能使用像 strtoul 这样将 32 位数字直接转换为二进制的函数.我必须使用公式.

mantissa[] is an int array where I have already converted the mantissa from a char array. When I get the value from formula, it has to be a binary number, and I have to convert it to decimal, so I will get the value of the number. Can you help me with storing the 23 bits of the mantissa? And, I mustn't use functions like strtoul that convert the 32-bit number directly into binary. I have to use formula.

推荐答案

鉴于所有的公式和示例编号以及一个计算器，以下代码的哪一部分难以正确?

Which part of the below code was hard to get right given all the formulas and sample numbers and a calculator?

#include <stdio.h>
#include <limits.h>

#if UINT_MAX >= 0xFFFFFFFF
typedef unsigned uint32;
#else
typedef unsigned long uint32;
#endif

#define C_ASSERT(expr) extern char CAssertExtern[(expr)?1:-1]

// Ensure uint32 is exactly 32-bit
C_ASSERT(sizeof(uint32) * CHAR_BIT == 32);

// Ensure float has the same number of bits as uint32, 32
C_ASSERT(sizeof(uint32) == sizeof(float));

double Ieee754SingleDigits2DoubleCheat(const char s[32])
{
  uint32 v;
  float f;
  unsigned i;
  char *p1 = (char*)&v, *p2 = (char*)&f;

  // Collect binary digits into an integer variable
  v = 0;
  for (i = 0; i < 32; i++)
    v = (v << 1) + (s[i] - '0');

  // Copy the bits from the integer variable to a float variable
  for (i = 0; i < sizeof(f); i++)
    *p2++ = *p1++;

  return f;
}

double Ieee754SingleDigits2DoubleNoCheat(const char s[32])
{
  double f;
  int sign, exp;
  uint32 mant;
  int i;

  // Do you really need strto*() here?
  sign = s[0] - '0';

  // Do you really need strto*() or pow() here?
  exp = 0;
  for (i = 1; i <= 8; i++)
    exp = exp * 2 + (s[i] - '0');

  // Remove the exponent bias
  exp -= 127;

  // Should really check for +/-Infinity and NaNs here

  if (exp > -127)
  {
    // Normal(ized) numbers
    mant = 1; // The implicit "1."
    // Account for "1." being in bit position 23 instead of bit position 0
    exp -= 23;
  }
  else
  {
    // Subnormal numbers
    mant = 0; // No implicit "1."
    exp = -126; // See your IEEE-54 formulas
    // Account for ".1" being in bit position 22 instead of bit position -1
    exp -= 23;
  }

  // Or do you really need strto*() or pow() here?
  for (i = 9; i <= 31; i++)
    mant = mant * 2 + (s[i] - '0');

  f = mant;

  // Do you really need pow() here?
  while (exp > 0)
    f *= 2, exp--;

  // Or here?
  while (exp < 0)
    f /= 2, exp++;

  if (sign)
    f = -f;

  return f;
}

int main(void)
{
  printf("%+g
", Ieee754SingleDigits2DoubleCheat("110000101100010010000000000000000"));
  printf("%+g
", Ieee754SingleDigits2DoubleNoCheat("010000101100010010000000000000000"));
  printf("%+g
", Ieee754SingleDigits2DoubleCheat("000000000100000000000000000000000"));
  printf("%+g
", Ieee754SingleDigits2DoubleNoCheat("100000000100000000000000000000000"));
  printf("%+g
", Ieee754SingleDigits2DoubleCheat("000000000000000000000000000000000"));
  printf("%+g
", Ieee754SingleDigits2DoubleNoCheat("000000000000000000000000000000000"));
  return 0;
}

输出(ideone):

-98.25
+98.25
+5.87747e-39
-5.87747e-39
+0
+0

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