Floating-Point Arithmetic

CS 301 Lecture, Dr. Lawlor

Ordinary integers can only represent integral values.  "Floating-point numbers" can represent non-integral values.   This is useful for engineering, science, statistics, graphics, and any time you need to represent numbers from the real world, which are rarely integral!

Floats store numbers in an odd way--they're really storing the number in scientific notation, like
    x = + 3.785746 * 105
Note that:
Scientific notation is designed to be compatible with slide rules (here's a circular slide rule demo); slide rules are basically a log table starting at 1.  This works because log(1) = 0, and log(a) + log(b) = log(ab).  But slide rules only give you the mantissa; you need to figure out the exponent yourself.  The "order of magnitude" guess that engineers (and I) like so much is just a calculation using zero significant digits--no mantissa, all exponent.

In binary, you can represent a non-integer like "two and three-eighths" as "10.011".  That is, there's:
for a total of two plus 1/4 plus 1/8, or "2+3/8".  Note that this is a natural measurement in carpenter's fractional inches, but it's a weird unnatural thing in metric-style decimal inches.  That is, fractions that are (negative) powers of two have a nice binary representation, but look weird in decimal (2 3/8 = 2.375); decimals have a nice decimal representation, but look very weird as a binary fraction.

Normalized Numbers

Scientific notation can represent the same number in several different ways:
    x = + 3.785746 * 105  = + 0.3785746 * 106 = + 0.003785746 * 107 = + 37.85746 * 104 

It's common to "normalize" a number in scientific notation so that:
  1. There's exactly one digit to the left of the decimal point.
  2. And that digit ain't zero.
This means the 105 version is the "normal" way to write the number above.

In binary, a "normalized" number *always* has a 1 at the left of the decimal point (if it ain't zero, it's gotta be one).  So sometimes there's no reason to even store the 1; you just know it's there!

(Note that there are also "denormalized" numbers, like 0.0, that don't have a leading 1.  This is how zero is represented--there's an implicit leading 1 only if the exponent field is nonzero, an implicit leading 0 if the exponent field is zero...)

Roundoff in Arithmetic

They're funny old things, floats.  The fraction part only stores so much precision; further bits are lost.  For example, in reality,
    1.2347654 * 104 = 1.234* 104 + 7.654* 100
But to three decimal places,
    1.234 * 104 = 1.234* 104 + 7.654* 100
which is to say, adding a tiny value to a great big value might not change the great big value at all, because the tiny value gets lost when rounding off to 3 places.  This "roundoff" has implications.

For example, adding one repeatedly will eventually stop doing anything:
float f=0.73;
while (1) {
volatile float g=f+1;
if (g==f) {
printf("f+1 == f at f=%.3f, or 2^%.3f\n",
return 0;
else f=g;
(executable NetRun link)
Recall that for integers, adding one repeatedly will *never* give you the same value--eventually the integer will wrap around, but it won't just stop moving like floats!

For another example, floating-point arithmetic isn't "associative"--if you change the order of operations, you change the result (up to roundoff):
    1.2355308 * 104 = 1.234* 104 + (7.654* 100 + 7.654* 100)
    1.2355308 * 104 = (1.234* 104 + 7.654* 100) + 7.654* 100
In other words, parenthesis don't matter if you're computing the exact result.  But to three decimal places,
    1.235 * 104 = 1.234* 104 + (7.654* 100 + 7.654* 100)
    1.234 * 104 = (1.234* 104 + 7.654* 100) + 7.654* 100
In the first line, the small values get added together, and together they're enough to move the big value.  But separately, they splat like bugs against the windshield of the big value, and don't affect it at all!
double lil=1.0;
double big=pow(2.0,64);
printf(" big+(lil+lil) -big = %.0f\n", big+(lil+lil) -big);
printf("(big+lil)+lil -big = %.0f\n",(big+lil)+lil -big);
(executable NetRun link)
float gnats=1.0;
volatile float windshield=1<<24;
float orig=windshield;
for (int i=0;i<1000;i++)
windshield += gnats;

if (windshield==orig) std::cout<<"You puny bugs can't harm me!\n";
else std::cout<<"Gnats added "<<windshield-orig<<" to the windshield\n";
(executable NetRun link)

In fact, if you've got a bunch of small values to add to a big value, it's more roundoff-friendly to add all the small values together first, then add them all to the big value:
float gnats=1.0;
volatile float windshield=1<<24;
float orig=windshield;
volatile float gnatcup=0.0;
for (int i=0;i<1000;i++)
gnatcup += gnats;
windshield+=gnatcup; /* add all gnats to the windshield at once */

if (windshield==orig) std::cout<<"You puny bugs can't harm me!\n";
else std::cout<<"Gnats added "<<windshield-orig<<" to the windshield\n";
(executable NetRun link)

Roundoff can be very annoying, but it doesn't matter if you don't care about exact answers, like in simulation (where "exact" means the same as the real world, which you'll never get anyway) or games.

One very frustrating fact is that roundoff depends on the precision you keep in your numbers.  This, in turn, depends on the size of the numbers.  For example, a "float" is just 4 bytes, but it's not very precise.  A "double" is 8 bytes, but it's more precise.  A "long double" is 12 bytes (or more!), but it's got tons of precision.

Roundoff in Representation

Sadly, 0.1 decimal is an infinitely repeating pattern in binary: 0.0(0011), with 0011 repeating.  This means multiplying by some *finite* pattern to approximate 0.1 is only an approximation of really dividing by the integer 10.0.  The exact difference is proportional to the precision of the numbers and the size of the input data:
for (int i=1;i<1000000000;i*=10) {
double mul01=i*0.1;
double div10=i/10.0;
double diff=mul01-div10;
std::cout<<"i="<<i<<" diff="<<diff<<"\n";
(executable NetRun link)

On the NetRun Pentium4 CPU, this gives:
i=1  diff=5.54976e-18
i=10 diff=5.55112e-17
i=100 diff=5.55112e-16
i=1000 diff=5.55112e-15
i=10000 diff=5.55112e-14
i=100000 diff=5.55112e-13
i=1000000 diff=5.54934e-12
i=10000000 diff=5.5536e-11
i=100000000 diff=5.54792e-10
Program complete. Return 0 (0x0)
That is, there's a factor of 10^-18 difference between double-precision 0.1 and the true 1/10!

Roundoff Taking Over Control

One place roundoff is very annoying is in your control structures.  For example, this loop will execute *seven* times, even though it looks like it should only execute *six* times:
for (double k=0.0;k<1.0;k+=1.0/6.0) {
printf("k=%a (about %.15f)\n",k,k);

(executable NetRun link)

The trouble is of course that 1/6 can't be represented exactly in floating-point, so if we add our approximation for 1/6 six times, we haven't quite hit 1.0, so the loop executes one additional time.  There are several possible fixes for this:
Any of these fixes will work, but you do have to realize this is a potential problem, and put the code in there!