How can I convert a float/double to ASCII without using sprintf or ftoa in C?

Question

How can I convert a float/double to ASCII without using sprintf or ftoa in C?

I am using an embedded system.

Answer 1

The approach you take will depend on the possible range of values. You certainly have some internal knowledge of the possible range, and you may only be interested in conversions within a more narrow range.

So, suppose you are only interested in the integer value. In this case, I would just assign the number to an int or long, at which point the problem becomes fairly obvious.

Or, suppose the range won't include any large exponents but you are interested in several digits of fraction. To get three digits of fraction, I might say int x = f * 1000;, convert x, and then insert the decimal point as a string operation.

Failing all of the above, a float or double has a sign bit, a fraction, and an exponent. There is a hidden 1 in the fraction. (The numbers are normalized until they have no leading zeroes, at which point they do one more shift to gain an extra bit of precision.) The number is then equal to the fraction (plus a leading '1') * 2 ** exponent. With essentially all systems using the IEEE 754 representation you can just use this Wikipedia IEEE 754 page to understand the format. It's not that different from just converting an integer.

For single precision, once you get the exponent and fraction, the value^{Note 1} of the number is then (frac / 2²³ + 1) * 2^exp, or frac * 2^{exp - 23} + 2^exp.

Here is an example that should get you started on a useful conversion:

$ cat t.c
#include <stdio.h>

void xconvert(unsigned frac)
{
  if (frac) {
    xconvert(frac / 10);
    printf("%c", frac % 10 | '0');
  }
}

void convert(unsigned i)
{
  unsigned sign, exp, frac;

  sign = i >> 31;
  exp  = (i >> (31 - 8)) - 127;
  frac = i & 0x007fffff;
  if (sign)
    printf("-");
  xconvert(frac);
  printf(" * 2 ** %d + 2 ** %d\n", exp - 23, exp);
  printf("\n");
}

int main(void)
{
  union {
     float f;
     unsigned i;
  } u;

  u.f = 1.234e9;
  convert(u.i);
  return 0;
}
$ ./a.out
1252017 * 2 ** 7 + 2 ** 30

---

^{Note 1. In this case the fraction is being converted as if the binary point was on the right instead of the left, with compensating adjustments then made to the exponent and hidden bit.}