how IEEE-754 floating point numbers work
floating-point, ieee-754, types
Solution
Think first in terms of decimal (base 10): 643.72 is:
- (6 * 102) +
- (4 * 101) +
- (3 * 100) +
- (7 * 10-1) +
- (2 * 10-2)
or 600 + 40 + 3 + 7/10 + 2/100.
That's because n0 is always 1, n-1 is the same as 1/n (for a specific case) and n-m is identical to 1/nm (for more general case).
Similarly, the binary number 1.1 is:
- (1 * 20) +
- (1 * 2-1)
with 20 being one and 2-1 being one-half.
In decimal, the numbers to the left of the decimal point have multipliers 1, 10, 100 and so on heading left from the decimal point, and 1/10, 1/100, 1/1000 heading right (i.e., 102, 101, 100, decimal point, 10-1, 10-2, ...).
In base-2, the numbers to the left of the binary point have multipliers 1, 2, 4, 8, 16 and so on heading left. The numbers to the right have multipliers 1/2, 1/4, 1/8 and so on heading right.
So, for example, the binary number:
101.00101
| | | |
| | | +- 1/32
| | +--- 1/8
| +------- 1
+--------- 4
is equivalent to:
4 + 1 + 1/8 + 1/32
or:
5
5 --
32
Problem
Let's say I have this: ``` float i = 1.5 ``` in binary, this float is represented as: 0 01111111 10000000000000000000000 I broke up the binary to represent the 'signed', 'exponent' and 'fraction' chunks. What I don't understand is how this represents 1.5. The exponent is 0 once you subtract the bias (127 - 127), and the fraction part with the implicit leading one is 1.1. How does 1.1 scaled by nothing = 1.5???