11

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???

Stephen Canon
  • 103,815
  • 19
  • 183
  • 269
Tony Stark
  • 24,588
  • 41
  • 96
  • 113

4 Answers4

28

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
paxdiablo
  • 854,327
  • 234
  • 1,573
  • 1,953
7

1.1 in binary is 1 + .5 = 1.5

pavpanchekha
  • 2,073
  • 1
  • 17
  • 23
1

The mantissa is essentially shifted by the exponent.

3 in binary is 0011
3>>1 in binary, equal to 3/2, is 0001.1
drawnonward
  • 53,459
  • 16
  • 107
  • 112
0

You want to read this - IEEE 754-1985

The actual standard is here

Romain Hippeau
  • 24,113
  • 5
  • 60
  • 79