Floating-Point Numbers

CS 301: Assembly Language Programming Lecture, Dr. Lawlor

Normalized Numbers

In C++, "float" and "double" store numbers in an odd way--internally they're really storing the number in scientific notation, like
x = + 3.785746 * 10⁵
Note that:

You only need one bit to represent the sign--plus or minus.
The exponent's just an integer, so you can store it as an integer.
The 3.785746 part, called the "mantissa" or just "fraction" part, can be stored as the integer 3785746 (at least as long as you can figure out where the decimal point goes!). Of course, we store the fraction part as a binary number.

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.

One problem is scientific notation can represent the same number in several different ways:
x = + 3.785746 * 10⁵ = + 0.3785746 * 10⁶ = + 0.03785746 * 10⁷ = + 37.85746 * 10⁴

It's common to "normalize" a number in scientific notation so that:

There's exactly one digit to the left of the decimal point.
And that digit is not zero.

This means the 10⁵ version above 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 there's no reason to even store the 1; you just know it's there!

Bits in a Floating-Point Number

Floats represent continuous values. But they do it using discrete bits.

A "float" (as defined by IEEE Standard 754) consists of three bitfields:

Sign	Exponent	Fraction (or "Mantissa")
1 bit-- 0 for positive 1 for negative	8 unsigned bits-- 127 means 2⁰ 137 means 2¹⁰	23 bits-- a binary fraction. Don't forget the implicit leading 1!

The sign is in the highest-order bit, the exponent in the next 8 bits, and the fraction in the remaining bits.

The hardware interprets a float as having the value:

value = (-1) ^sign * 2 ^{(exponent-127)}* 1.fraction

Note that the mantissa has an implicit leading binary 1 applied. The 1 isn't stored, which actually causes some headaches. (Even worse, if the exponent field is zero, then it's an implicit leading 0; a "denormalized" number as we'll talk about on Wednesday.)

For example, the value "8" would be stored with sign bit 0, exponent 130 (==3+127), and mantissa 000... (without the leading 1), since:

8 = (-1) ⁰ * 2 ^(130-127)* 1.0000....

You can stare at the bits inside a float by converting it to an integer. The quick and dirty way to do this is via a pointer typecast, but modern compilers will sometimes over-optimize this, especially in inlined code:

void print_bits(float f) {
	int i=*reinterpret_cast<int *>(&f); /* read bits with "pointer shuffle" */
	std::cout<<" float "<<std::setw(10)<<f<<" = ";
	for (int bit=31;bit>=0;bit--) {
		if (i&(1<<bit)) std::cout<<"1"; else std::cout<<"0";
		if (bit==31) std::cout<<" ";
		if (bit==23) std::cout<<" (implicit 1).";
	}
	std::cout<<std::endl;
}

int foo(void) {
	print_bits(0.0);
	print_bits(-1.0);
	print_bits(1.0);
	print_bits(2.0);
	print_bits(4.0);
	print_bits(8.0);
	print_bits(1.125);
	print_bits(1.25);
	print_bits(1.5);
	print_bits(1+1.0/10);
	return sizeof(float);
}

(Try this in NetRun now!)

The official way to dissect the parts of a float is using a "union" and a bitfield like so:

/* IEEE floating-point number's bits:  sign  exponent   mantissa */
struct float_bits {
	unsigned int fraction:23; /**< Value is binary 1.fraction ("mantissa") */
	unsigned int exp:8; /**< Value is 2^(exp-127) */
	unsigned int sign:1; /**< 0 for positive, 1 for negative */
};

/* A union is a struct where all the fields *overlap* each other */
union float_dissector {
	float f;
	float_bits b;
};

float_dissector s;
s.f=8.0;
std::cout<<s.f<<"= sign "<<s.b.sign<<" exp "<<s.b.exp<<"  fract "<<s.b.fraction<<"\n";
return 0;

(Executable NetRun link)

I like to joke that a union misused to convert bits between incompatible types is an "unholy union".

In addition to the 32-bit "float", there are several other different sizes of floating-point types:

C Datatype	Size	Approx. Precision	Approx. Range	Exponent Bits	Fraction Bits	+-1 range
float	4 bytes (everywhere)	1.0x10^-7	10³⁸	8	23	2²⁴
double	8 bytes (everywhere)	2.0x10^-15	10³⁰⁸	11	52	2⁵³
long double	12-16 bytes (if it even exists)	2.0x10^-20	10⁴⁹³²	15	64	2⁶⁵
half float	2 bytes (only on GPUs)	1.0x10^-3	10⁵	5	10	2¹¹

Nowadays floats have roughly the same performance as integers: addition, subtraction, or multiplication all take about a nanosecond. That is, floats are now cheap, and you can consider using floats for all sorts of stuff--even when you don't care about fractions! The advantages of using floats are:

Floats can store fractional numbers.
Floats never overflow; they hit "infinity" instead.
"double" has more bits than "int" (but less than "long").

Due to these advantages, many interpreted languages including JavaScript have only one numeric type, usually double-precision float.

Roundoff in Arithmetic

They're funny old things, floats. The fraction part (mantissa) only stores so much precision; further bits are lost. For example, in reality,
    1.234* 10⁴ + 7.654* 10⁰= 1.2347654 * 10⁴
But if we only keep three decimal places,
    1.234* 10⁴ + 7.654* 10⁰ = 1.234 * 10⁴
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.   To avoid this "roundoff error", when you're doing arithmetic by hand, people recommend keeping lots of digits, and only rounding once, at the end. But for a given value of "lots of digits", did you keep enough?

For example, on a real computer adding one to a float repeatedly will eventually stop changing the float!

float f=0.73;
while (1) {
	volatile float g=f+1;
	if (g==f) {
		std::cout<<"f+1 == f  at f="<< f <<", or 2^"<< log(f)/log(2.0) <<std::endl;
		return 0;
	}
	else f=g;
}

(Try this in NetRun now!)

For "double", you can add one more times, but eventually the double will stop changing despite your additions. 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!

This has really weird effects. For example, floating-point arithmetic isn't "associative"--if you change the order of operations, you change the result due to accumulated roundoff. In exact arithmetic:
    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(2.0,53); //<- carefully chosen for IEEE 64-bit float (52 bits of fraction + implicit 1)
std::cout<<" big+(lil+lil) -big = "<<  big+(lil+lil) -big <<std::endl;
std::cout<<"(big+lil)+lil  -big = "<< (big+lil)+lil  -big <<std::endl;

(Try this in NetRun now!)

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's not the end of the world if you don't care about exact answers, like in computer games, and even in many simulations (where "exact" is unmeasureable anyway). You just need to be able to estimate the amount of roundoff, and make sure it's not too much.

However, the amount of 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, so it's not very precise. A "double" is 8 bytes, and so more precise. A "long double" is 12 bytes (or more!), using more memory, but it's got tons of precision. There's often a serious tradeoff between precision and space (and time), so just using long double for everything isn't a good idea: your program may get bigger and slower, and you still might not have enough 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)

In a perfect world, multiplying by 0.1 and dividing by 10 would give the exact same result. But in reality, 0.1 has to be approximated by a finite series of binary digits, while the integer 10 can be stored exactly, so on NetRun's Pentium4 CPU, this code 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! This can add up over time.