How to manually parse a floating point number from a string

Question

Of course most languages have library functions for this, but suppose I want to do it myself.

Suppose that the float is given like in a C or Java program (except for the 'f' or 'd' suffix), for example "4.2e1", ".42e2" or simply "42". In general, we have the "integer part" before the decimal point, the "fractional part" after the decimal point, and the "exponent". All three are integers.

It is easy to find and process the individual digits, but how do you compose them into a value of type float or double without losing precision?

I'm thinking of multiplying the integer part with 10^n, where n is the number of digits in the fractional part, and then adding the fractional part to the integer part and subtracting n from the exponent. This effectively turns 4.2e1 into 42e0, for example. Then I could use the pow function to compute 10^exponent and multiply the result with the new integer part. The question is, does this method guarantee maximum precision throughout?

Any thoughts on this?

Solution 1

I would directly assemble the floating point number using its binary representation.

Read in the number one character after another and first find all digits. Do that in integer arithmetic. Also keep track of the decimal point and the exponent. This one will be important later.

Now you can assemble your floating point number. The first thing to do is to scan the integer representation of the digits for the first set one-bit (highest to lowest).

The bits immediately following the first one-bit are your mantissa.

Getting the exponent isn't hard either. You know the first one-bit position, the position of the decimal point and the optional exponent from the scientific notation. Combine them and add the floating point exponent bias (I think it's 127, but check some reference please).

This exponent should be somewhere in the range of 0 to 255. If it's larger or smaller you have a positive or negative infinite number (special case).

Store the exponent as it into the bits 24 to 30 of your float.

The most significant bit is simply the sign. One means negative, zero means positive.

It's harder to describe than it really is, try to decompose a floating point number and take a look at the exponent and mantissa and you'll see how easy it really is.

Btw - doing the arithmetic in floating point itself is a bad idea because you will always force your mantissa to be truncated to 23 significant bits. You won't get a exact representation that way.

Solution 2

All of the other answers have missed how hard it is to do this properly. You can do a first cut approach at this which is accurate to a certain extent, but until you take into account IEEE rounding modes (et al), you will never have the right answer. I've written naive implementations before with a rather large amount of error.

If you're not scared of math, I highly recommend reading the following article by David Goldberg, What Every Computer Scientist Should Know About Floating-Point Arithmetic. You'll get a better understanding for what is going on under the hood, and why the bits are laid out as such.

My best advice is to start with a working atoi implementation, and move out from there. You'll rapidly find you're missing things, but a few looks at strtod's source and you'll be on the right path (which is a long, long path). Eventually you'll praise insert diety here that there are standard libraries.

/* use this to start your atof implementation */

/* atoi - [email protected] */
/* PUBLIC DOMAIN */
long atoi(const char *value) {
  unsigned long ival = 0, c, n = 1, i = 0, oval;
  for( ; c = value[i]; ++i) /* chomp leading spaces */
    if(!isspace(c)) break;
  if(c == '-' || c == '+') { /* chomp sign */
    n = (c != '-' ? n : -1);
    i++;
  }
  while(c = value[i++]) { /* parse number */
    if(!isdigit(c)) return 0;
    ival = (ival * 10) + (c - '0'); /* mult/accum */
    if((n > 0 && ival > LONG_MAX)
    || (n < 0 && ival > (LONG_MAX + 1UL))) {
      /* report overflow/underflow */
      errno = ERANGE;
      return (n > 0 ? LONG_MAX : LONG_MIN);
    }
  }
  return (n>0 ? (long)ival : -(long)ival);
}

Solution 3

The "standard" algorithm for converting a decimal number to the best floating-point approximation is William Clinger's How to read floating point numbers accurately, downloadable from here. Note that doing this correctly requires multiple-precision integers, at least a certain percentage of the time, in order to handle corner cases.

Algorithms for going the other way, printing the best decimal number from a floating-number, are found in Burger and Dybvig's Printing Floating-Point Numbers Quickly and Accurately, downloadable here. This also requires multiple-precision integer arithmetic

See also David M Gay's Correctly Rounded Binary-Decimal and Decimal-Binary Conversions for algorithms going both ways.

Solution 4

You could ignore the decimal when parsing (except for its location). Say the input was: 156.7834e10... This could easily be parsed into the integer 1567834 followed by e10, which you'd then modify to e6, since the decimal was 4 digits from the end of the "numeral" portion of the float.

Precision is an issue. You'll need to check the IEEE spec of the language you're using. If the number of bits in the Mantissa (or Fraction) is larger than the number of bits in your Integer type, then you'll possibly lose precision when someone types in a number such as:

5123.123123e0 - converts to 5123123123 in our method, which does NOT fit in an Integer, but the bits for 5.123123123 may fit in the mantissa of the float spec.

Of course, you could use a method that takes each digit in front of the decimal, multiplies the current total (in a float) by 10, then adds the new digit. For digits after the decimal, multiply the digit by a growing power of 10 before adding to the current total. This method seems to beg the question of why you're doing this at all, however, as it requires the use of the floating point primitive without using the readily available parsing libraries.

Anyway, good luck!

Solution 5

Yes, you can decompose the construction into floating point operations as long as these operations are EXACT, and you can afford a single final inexact operation.

Unfortunately, floating point operations soon become inexact, when you exceed precision of mantissa, the results are rounded. Once a rounding "error" is introduced, it will be cumulated in further operations...
So, generally, NO, you can't use such naive algorithm to convert arbitrary decimals, this may lead to an incorrectly rounded number, off by several ulp of the correct one, like others have already told you.

BUT LET'S SEE HOW FAR WE CAN GO:

If you carefully reconstruct the float like this:

if(biasedExponent >= 0)
    return integerMantissa * (10^biasedExponent);
else
    return integerMantissa / (10^(-biasedExponent));

there is a risk to exceed precision both when cumulating the integerMantissa if it has many digits, and when raising 10 to the power of biasedExponent...

Fortunately, if first two operations are exact, then you can afford a final inexact operation * or /, thanks to IEEE properties, the result will be rounded correctly.

Let's apply this to single precision floats which have a precision of 24 bits.

10^8 > 2^24 > 10^7

Noting that multiple of 2 will only increase the exponent and leave the mantissa unchanged, we only have to deal with powers of 5 for exponentiation of 10:

5^11 > 2^24 > 5^10

Though, you can afford 7 digits of precision in the integerMantissa and a biasedExponent between -10 and 10.

In double precision, 53 bits,

10^16 > 2^53 > 10^15
5^23 > 2^53 > 5^22

So you can afford 15 decimal digits, and a biased exponent between -22 and 22.

It's up to you to see if your numbers will always fall in the correct range... (If you are really tricky, you could arrange to balance mantissa and exponent by inserting/removing trailing zeroes).

Otherwise, you'll have to use some extended precision.
If your language provides arbitrary precision integers, then it's a bit tricky to get it right, but not that difficult, I did this in Smalltalk and blogged about it at http://smallissimo.blogspot.fr/2011/09/clarifying-and-optimizing.html and http://smallissimo.blogspot.fr/2011/09/reviewing-fraction-asfloat.html

Note that these are simple and naive implementations. Fortunately, libc is more optimized.

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