Given a double, need to find how many digits in total

Question

I have a double which is not necessarily positive but usually. It can be 0.xxxx000 or X.xxxx00000 or XX.00000 or 0.xxx0xxx00000, where eventually there are all 0's to the right of the last number. I need to keep track of how many digits there are. I've been having trouble with this, any help? This is C.

Answer 1

Use sprintf to turn it into a string and do whatever counting/testing you need to do on the digits

Answer 2

A double has 52 mantissa bits plus an implicit "1" bit, so you should be able to type-pun a double pointer to a 64-bit integer (getting the raw bits into an integer), &= this with (1<<52)-1, and |= the result with (1<<52).

The log10 of that would be the number of decimal digits.

Though, I'm almost inclined to say "go with jonsca's solution" because it is so ingeniously simple (it deserves a +1 in any case for being KISS).

Answer 3

The representation of the double is not decimal - it is binary (like all the other numbers in a computer). The problem you defined makes little sense really. Consider the example: number 1.2 is converted to binary - 1+1/5 = 1.(0011) binary [0011 in period]. If you cut it to 52 bits of precision (double) - you'll have 1.0011001100110011001100110011001100110011001100110011 binary that equals 1+(1-1/2^52)/5. If you represent this number in decimal form precisely you will get 52 decimals before all-zeroes that is a lot more than the maximum decimal precision of a double that is 16 digits (and all those digits of representation from 17 to 52 are just meaningless).

Anyway if you have purely abstract problem (like in school):

int f( double x )
{
  int n = 0;

  x = fabs(x);
  x -= floor(x);

  while( x != floor(x) )
  {
    x *= 2;
    ++n;
  }

  return n;
}

The function returns number of binary digits before all-zeroes and it is also the number of decimal digits before all-zeroes (the last decimal digit is always 5 if returned value > 0).