User:Berni44/Floatingpoint

An introduction to floating point numbers

You probably already know, that strange things can happen, when using floating point numbers.

An example:

import std.stdio;

void main()
{
    float a = 1000;
    float b = 1/a;
    float c = 1/b;
    writeln("Is ",a," == ",c,"? ",a==c?"Yes!":"No!");
}

Did you guess the answer?

Is 1000 == 1000? No!

To understand this strange behavior we have to look at the bit representation of the numbers involved. Unfortunately, floats have already 32 bits and with that many 0s and 1s it can easily happen that one can't see the forest for the trees.

For that reason I'll start with smaller floating point numbers, that I call nano floats. Nano floats have only 6 bits.

Nano floats

Floating point numbers consist of three parts: A sign bit, which is always exactly one bit, an exponent and a mantissa. Nano floats use 3 bits for the exponent and 2 bits for the mantissa. For example 1 100 01 is the bit representation of a nano float. Which number does this bit pattern represent?

You can think of floating point numbers as numbers written in scientific notation, known from physics: For example the speed of light is about +2.9979 * 10^^8 m/s. Here we've got the sign bit (+), the exponent (8) and the mantissa (2.9979). Putting this together, we could write that number as + 8 2.9979. This looks already a little bit like our number 1 100 01.

What we still need, is to know how the parts of that number are decoded. Let's start with the sign bit, which is easy. A 0 is + and a 1 is −. We now know, that our number is negative.

Next the exponent: 100 is the binary code of 4. So our exponent is 4? No, it's not that easy. Exponents can also be negative. To achieve this, we have to subtract the so called bias. The bias can be calculated from the number of bits of the exponent. If r is the number of bits of the exponent, the bias is 2^^(r−1)−1. Here, we've got r=3, and therefore the bias is 2^^2−1=3, and finally we get our exponent, it's 4−3=1.

Now the mantissa. We've seen in the speed of light example above, that the mantissa was 2.9979. Note, that it is usual for scientific notation, that there is always exactly one integral digit in the mantissa, in this case 2. Additionally there are four fractional digits: 9979. Now, floating point numbers use binary code instead of decimal code. This implies, that the integral digit is (almost, see below) always 1. It would be a waste to save this 1 in our number. Therefore it's omitted. Adding it to our mantissa, we've got 1.01 in binary code, which is 1.25 in decimal code.

Putting all together we have: 1 100 01 = − 1.25 * 2 ^^ 1 = −2.5.

Exercise

I'll add exercises throughout this document. I recommend to do them — you'll acquire a much better feeling for floating point numbers, when you do this on your own, instead of peeking at the answers. But of course, it's up to you.

Exercise 1: Write down all 64 bit patterns of nano floats in a table and calculate the value, which is represented by that value:

Bit pattern	Value
0 000 00
0 000 01
0 000 10
0 000 11
0 001 00
...
1 100 01	−2.5
...
1 111 11

Zero and denormalized numbers

The table from the exercise above can be visualized on a number line:

One clearly sees, that on the outside, the numbers are sparse. The count of numbers increases, while approaching 0 from both sides. But then, suddenly they stop and leave a gap at zero. Let's zoom in:

Now we can clearly see the gap: There is no 0. The reason for this is, that 0 is the only number, where the integral bit of the mantissa has to be 0, because there is no 1 bit available.

To get around this, numbers with exponent 000, which are called denormalized numbers or subnormal numbers, are treated special: These numbers are considered to have an integral 0 bit implied to the mantissa and the exponent is increased by one. So 1 000 10 would be decoded as - 0.5 * 2^(-2) = -0.125.

And when the mantissa is 00 we've got our zero: 0 000 00 = +0. Unfortunately, there is a second zero: 1 000 00 = −0.

Infinity and Not a Number

There is an other exponent, which is treated special: 111. This time, if the mantissa is 00 it is considered to be infinity and if it is 11 it denotes a number, that is not a number, a so called NaN. Other values for the mantissa are also considered to be NaNs, but this time with some special meanings attached, which is beyond the scope of this article. And yes, there are also the minus versions of all of these NaNs.

With that, our number line looks like this:

And the zoomed in version looks like this:

It can now clearly be seen, that the center is equispaced and the outsides have been truncated somewhat.

Exercise

Exercise 2: Add a third column to the table from exercise 1 and write the special values in that column.

Back to our example from the beginning

Now we can track down the problems in the example at the beginning somewhat. Nanofloats cannot show 1000. But with 1.75, which can be displayed as a nanofloat, we have a similar calculation:

nanofloat a = 1.75;
nanofloat b = 1/a;
nanofloat c = 1/a;

This time we can use our tables from the exercises to look up the results:

1/1.75 = 0.57142857..., which cannot exactly be coded with a nanofloat. We have to choose between 0.5 and 0.625. Normally, floating point units are supposed to round to the nearest possibility in such cases. In this case 0.625 is closer to 0.57142857... than 0.5; that is, b = 0.625.

1/0.625 = 1.6. Again a value, that cannot exactly be coded as nanofloat. We've got 1.5 and 1.75 as a choice. 1.5 is closer to 1.6 than 1.75 and therefore c = 1.5. We end up with a != c.

But what would <syntaxhighlight lang=D>

   writeln("Is ",a," == ",c,"? ",a==c?"Yes!":"No!");

</syntaxhightlight> produce here?

Well, it would be

Is 1.75 == 1.5? No!

Now you may ask, why the example with the floats produced 1000 == 1000 and did not display two different values.

The answer to this question can be found, when looking into the implementation of writeln: You'll find a call to formattedWrite which can be found in std.format, with a format specifier of %s. For floating point numbers, %s is treated identical to %g, which has some design flaws. In our case, %g rounds too eagerly: The correct value of c is 999.99993896484375. But the default output of %g is limited to 6 significant digits, which means, for the sake of writing the number, it is rounded to 1000.

... to be continued

Solutions

Exercise 1 and 2:

Bit pattern	Value	Special value
0 000 00	0.125	0.0
0 000 01	0.15625	0.0625
0 000 10	0.1875	0.125
0 000 11	0.21875	0.1875
0 001 00	0.25
0 001 01	0.3125
0 001 10	0.375
0 001 11	0.4375
0 010 00	0.5
0 010 01	0.625
0 010 10	0.75
0 010 11	0.875
0 011 00	1
0 011 01	1.25
0 011 10	1.5
0 011 11	1.75
0 100 00	2
0 100 01	2.5
0 100 10	3
0 100 11	3.5
0 101 00	4
0 101 01	5
0 101 10	6
0 101 11	7
0 110 00	8
0 110 01	10
0 110 10	12
0 110 11	14
0 111 00	16	infinity
0 111 01	20	special NaN
0 111 10	24	special NaN
0 111 11	28	NaN
1 000 00	−0.125	−0,0
1 000 01	−0.15625	−0.0625
1 000 10	−0.1875	−0.125
1 000 11	−0.21875	−0.1875
1 001 00	−0.25
1 001 01	−0.3125
1 001 10	−0.375
1 001 11	−0.4375
1 010 00	−0.5
1 010 01	−0.625
1 010 10	−0.75
1 010 11	−0.875
1 011 00	−1
1 011 01	−1.25
1 011 10	−1.5
1 011 11	−1.75
1 100 00	−2
1 100 01	−2.5
1 100 10	−3
1 100 11	−3.5
1 101 00	−4
1 101 01	−5
1 101 10	−6
1 101 11	−7
1 110 00	−8
1 110 01	−10
1 110 10	−12
1 110 11	−14
1 111 00	−16	−infinity
1 111 01	−20	−special NaN
1 111 10	−24	−special NaN
1 111 11	−28	−NaN