# IEEE-754 Floating Point Numbers

• Published

Many programmers are aware of floating point numbers, typically called floats, and their value for storing decimal values. Most of them are also aware of some vague danger in using them which can cause some really bizarre-seeming results.

For example, in Ruby,

``````0.1 + 0.2 = 0.30000000000000004
``````

This is due to the way we store these decimal values.

Floats use scientific notation in binary format, meaning they use 2 as a base instead of 10. These rounding errors occur because we can’t quite represent the exact correct answer within our limits of efficient time or memory usage.

Floats are either stored as 32 or 64 bits.

1. 1 sign bit

2. 8 or 11 exponent bits

3. 23 or 52 mantissa bits

“Mantissa” is the word for the number between being multiplied in scientific notation.

There are several caveats to know up front:

1. The exponent is unsigned, but biased by 127 or 1023 depending on 32 or 64 bit. This means For minimal exponent, use value 1. For value of 0, use the value of the offset. For maximal exponent, use double the value of the offset.

2. Mantissa bits are read left to right, contrary to typical binary, and represent fractional binary values.

3. The mantissa bits should exclude a 1 to the left. For correct scientific notation, we should always have a 1 in the most significant value. Since it’s always the same value, we skip it when encoding.

For example, having 0.011 wouldn’t make sense, so we left shift until there is a 1 and adjust the exponent.

## A Few Examples

The following are a few examples of using floating point encoding by hand in order to better understand what’s happening under the hood.

### Encoding Example

Let’s encode the following number to floating point.

``````-450.875
``````
1. This number is negative, so it’s sign bit is one.

`````` S = 1
``````
2. Encode the integral part of the number.

`````` 450 = 0b0001 1100 0010
``````
3. Find the value of the fractional part of the number.

`````` 0.875 x 2 = 1.75   
0.75 x 2  = 1.5    
0.5 x 2   = 1.0    
0.0 x 2   = 0.0    

0.875 = 0b1110
= (1/2) + (1/4) + (1/8)
``````
4. Our Mantissa bits are the combination of integral and fractional parts.

`````` M = 0b0001 1100 0010.1110
``````
5. The exponent is the number of spaces we need to move the decimal to be behind the leading one, plus the bias.

`````` M = 1.11000010111
^        ^ 8 spaces moved

Exponent = 8 + 127 = 10000111
``````
6. Combine all parts S + E + M

`````` 1 10000111 11000010111000000000000
Or, in hex 0xc3e17000
``````

Let’s add together two floating point values after they are encoded.

``````25.125 + -122.625
``````
1. Convert both to floating point

`````` 0 10000011 10010010000000000000000
1 10000101 11101010100000000000000
``````
2. Match the exponents by right shifting the mantissa bits of the smaller number

`````` 0 10000101 01100100100000000000000
1 10000101 11101010100000000000000
``````

Rember to incorporate the leading 1 before the bits began.

``````  01100100100000000000000
-11101010100000000000000
––––––––––––––––––––––––
-10000110000000000000000
``````
4. The resulting bits were negative so the result is negative. The exponent is what was matched.

`````` 1 10000101 10000110000000000000000
``````
5. Just to verify the answer is correct, convert to decimal

`````` 110001.1 = -97.5
``````
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