9 KiB
Floating Point
Floating point arithmetic (normally just float) is a method of computer representation of fractional numbers, i.e. numbers with higher than integer precision (such as 5.13), which is more complex than e.g. fixed point. The core idea of it is to use a radix point that's not fixed but can move around so as to allow representation of both very small and very big values. Nowadays floating point is the standard way of approximating real numbers in computers, basically all of the popular programming languages have a floating point data type that adheres to the IEEE 754 standard, all personal computers also have the floating point hardware unit (FPU) and so it is widely used in all modern programs. However most of the time a simpler representation of fractional numbers, such as the mentioned fixed point, suffices, and weaker computers (e.g. embedded) may lack the hardware support so floating point operations are emulated in software and therefore slow -- for these reasons we consider floating point bloat and recommend the preference of fixed point.
Is floating point literal evil? Well, of course not, but it is extremely overused. You may need it for precise scientific simulations, e.g. numerical integration, but as our small3dlib shows, you can comfortably do even 3D rendering without it. So always consider whether you REALLY need float.
How It Works
Floats represent numbers by representing two main parts: the base -- actual encoded digits, called mantissa (or significand etc.) -- and the position of the radix point. The position of radix point is called the exponent because mathematically the floating point works similarly to the scientific notation of extreme numbers that use exponentiation. For example instead of writing 0.0000123 scientists write 123 * 10^-7 -- here 123 would be the mantissa and -7 the exponent.
Though various numeric bases can be used, in computers we normally use base 2, so let's consider it from now on. So our numbers will be of format:
mantissa * 2^exponent
Note that besides mantissa and exponent there may also be other parts, typically there is also a sign bit that says whether the number is positive or negative.
Let's now consider an extremely simple floating point format based on the above. Keep in mind this is an EXTREMELY NAIVE inefficient format that wastes values. We won't consider negative numbers. We will use 6 bits for our numbers:
- 3 leftmost bits for mantissa: This allows us to represent 2^3 = 8 base values: 0 to 7 (including both).
- 3 rightmost bits for exponent: We will encode exponent in two's complement so that it can represent values from -4 to 3 (including both).
So for example the binary representation 110011 stores mantissa 110 (6) and exponent 011 (3), so the number it represents is 6 * 2^3 = 48. Similarly 001101 represents 1 * 2^-3 = 1/8 = 0.125.
Note a few things: firstly our format is shit because some numbers have multiple representations, e.g. 0 can be represented as 000000, 000001, 000010, 000011 etc., in fact we have 8 zeros! That's unforgivable and formats used in practice address this (usually by prepending an implicit 1 to mantissa).
Secondly notice the non-uniform distribution of our numbers: while we have a nice resolution close to 0 (we can represent 1/16, 2/16, 3/16, ...) but low resolution in higher numbers (the highest number we can represent is 56 but the second highest is 48, we can NOT represent e.g. 50 exactly). Realize that obviously with 6 bits we can still represent only 64 numbers at most! So float is NOT a magical way to get more numbers, with integers on 6 bits we can represent numbers from 0 to 63 spaced exactly by 1 and with our floating point we can represent numbers spaced as close as 1/16th but only in the region near 0, we pay the price of having big gaps in higher numbers.
Also notice that thing like simple addition of numbers become more difficult and time consuming, you have to include conversions and rounding -- while with fixed point addition is a single machine instruction, same as integer addition, here with software implementation we might end up with dozens of instructions (specialized hardware can perform addition fast but still, not all computer have that hardware).
Rounding errors will appear and accumulate during computations: imagine the operation 48 + 1/8. Both numbers can be represented in our system but not the result (48.125). We have to round the result and end up with 48 again. Imagine you perform 64 such additions in succession (e.g. in a loop): mathematically the result should be 48 + 64 * 1/8 = 56, which is a result we can represent in our system, but we will nevertheless get the wrong result (48) due to rounding errors in each addition. So the behavior of float can be non intuitive and dangerous, at least for those who don't know how it works.
Standard Float Format: IEEE 754
IEEE 754 is THE standard that basically all computers use for floating point nowadays -- it specifies the exact representation of floating point numbers as well as rounding rules, required operations applications should implement etc. However note that the standard is kind of shitty -- even if we want to use floating point numbers there exist better ways such as posits that outperform this standard. Nevertheless IEEE 754 has been established in the industry to the point that it's unlikely to go anytime soon. So it's good to know how it works.
Numbers in this standard are signed, have positive and negative zero (oops), can represent plus and minus infinity and different NaNs (not a number). In fact there are thousands to billions of different NaNs which are basically wasted values. These inefficiencies are addressed by the mentioned posits.
Briefly the representation is following (hold on to your chair): leftmost bit is the sign bit, then exponent follows (the number of bits depends on the specific format), the rest of bits is mantissa. In mantissa implicit 1. is considered (except when exponent is all 0s), i.e. we "imagine" 1. in front of the mantissa bits but this 1 is not physically stored. Exponent is in so called biased format, i.e. we have to subtract half (rounded down) of the maximum possible value to get the real value (e.g. if we have 8 bits for exponent and the directly stored value is 120, we have to subtract 255 / 2 = 127 to get the real exponent value, in this case we get -7). However two values of exponent have special meaning; all 0s signify so called denormalized (also subnormal) number in which we consider exponent to be that which is otherwise lowest possible (e.g. -126 in case of 8 bit exponent) but we do NOT consider the implicit 1 in front of mantissa (we instead consider 0.), i.e. this allows storing zero (positive and negative) and very small numbers. All 1s in exponent signify either infinity (positive and negative) in case mantissa is all 0s, or a NaN otherwise -- considering here we have the whole mantissa plus sign bit unused, we actually have many different NaNs (WTF), but usually we only distinguish two kinds of NaNs: quiet (qNaN) and signaling (sNaN, throws and exception) that are distinguished by the leftmost bit in mantissa (1 for qNaN, 0 for sNaN).
The standard specifies many formats that are either binary or decimal and use various numbers of bits. The most relevant ones are the following:
| name | M bits | E bits | smallest and biggest number | precision <= 1 up to |
|---|---|---|---|---|
| binary16 (half precision) | 10 | 5 | 2^(-24), 65504 | 2048 |
| binary32 (single precision, float) | 23 | 8 | 2^(-149), 2127 * (2 - 2^-23) ~= 3 * 10^38 | 16777216 |
| binary64 (double precision, double) | 52 | 11 | 2^(-1074), ~10^308 | 9007199254740992 |
| binary128 (quadruple precision) | 112 | 15 | 2^(-16494), ~10^4932 | ~10^34 |
Example? Let's say we have float (binary34) value 11000000111100000000000000000000: first bit (sign) is 1 so the number is negative. Then we have 8 bits of exponent: 10000001 (129) which converted from the biased format (subtracting 127) gives exponent value of 2. Then mantissa bits follow: 11100000000000000000000. As we're dealing with a normal number (exponent bits are neither all 1s nor all 0s), we have to imagine the implicit 1. in front of mantissa, i.e. our actual mantissa is 1.11100000000000000000000 = 1.875. The final number is therefore -1 * 1.875 * 2^2 = -7.5.