# Program 20: Here to There

## Floating point numbers #

How does a floating point number actually work? IEEE 754 is the tech sheet that talks about floating point numbers. For single precision, it says that the layout is:

Sign (+/-) | Exponent | Fraction |
---|---|---|

bit 31 | bits 23-30 | bit 0-22 |

For double precision it is:

Sign (+/-) | Exponent | Fraction |
---|---|---|

bit 63 | bits 52-62 | bit 0-51 |

The sign bit is the easy part. If the bit is 1, it is negative and if it’s 0 the number is positive.

The next set of bits are the exponent for the number 2. When representing the exponent, the number here will never be read as negative. Instead, the exponent is read as a positive and a bias is removed. (127 for single and 1023 for double) So if, in a single precision float, the exponent is written as 129, the exponent is really 2. (129-127)

The fraction or mantissa is represented as 1 (not encoded) plus the addition
of 2^-x where x is the 1-indexed bit from the most significant side.
If the fraction started with `101011`

… it would represent.

If you really want to dig into this conversion, this website is one of the most clear and concise explanations I’ve found.

## Program 20: Here to There Video - Coming Soon

**Note**: What I really really really wanted to do here was have you write a float to
ASCII implementation. As there are multiple papers written on the subject and it’s
quite complex, I wanted to find another exercise that would allow you to interact
with a floating point number. If you have a better exercise for this, please let
me know.

For this exercise, you are given two points. Find the distance between the two points and then output the parts of the equation. If the the below bit representation was the result:

Sign (+/-) | Exponent | Fraction |
---|---|---|

0 | 11110011 | 11100000000000000000000 |

+ | 243 | 1/2 + 1/4 + 1/8 |

You would output:

```
Sign: +
Exponent: 116
Fraction (1/x): 1 2 4 8
```

Input Points: (4, 5), (2, 1)