Download The IEEE 754 standard floating point number system 32 bit floating

The IEEE 754 standard floating point number system
32 bit floating point. A floating point number has the form
±(1.b1 b2 . . . b23 )2 × 2M .
The sign is determined by the sign bit:
(
± = (−1)s =
1, s = 0
−1, s = 1
The exponent M is determined by 8 exponent bits e1 e2 . . . e8 as follows. The cases of all
zeros and all ones are used to represent special values (zero, positive infinity, NaN, etc.).
Otherwise we interpret e1 e2 . . . e8 as a binary integer, subtract the bias value 127, and the
result is M . Thus the least exponent is M = −126 with the bit string e1 e2 . . . e8 = 00000001,
and the greatest possible exponent is M = 127 with bit string e1 e2 . . . e8 = 11111110.
The remaining 23 mantissa bits and the implied leading 1 then determine the fractional
part:
(1.b1 b2 . . . b23 )2 = 1 + b1 /2 + b2 /4 + . . . + b23 /223 .
The least value, when all the mantissa bits are 0, is 1, and the greatest value, when all the
mantissa bits are 1, is 2 − 223 .
Combining these we see that the largest floating point number has the bit string
0 11111110 11111111111111111111111
which represents the value (2 − 223 ) × 2127 = 2128 − 2104 ≈ 3.4 × 1038 .
The smallest positive floating point number has bit string
0 00000001 00000000000000000000000
with the value 2−126 ≈ 1.2 × 10−38 .
The machine epsilon, which is the spacing between consecutive floating point numbers
in the interval [1, 2], is 2−23 ≈ 1.2 × 10−7 .
smallest exponent
−126
largest exponent
+127
smallest positive number 1.2 × 10−38
largest number
3.4 × 10+38
machine epsilon
1.2 × 10−7
64 bit floating point
This uses 1 sign bit, 11 exponent bits, and 53 bits for the mantissa, not counting the
leading 1. The bias is 1023. This gives:
smallest exponent
−1022
largest exponent
+1023
smallest positive number 2.2 × 10−308
largest number
1.8 × 10+308
machine epsilon
1.1 × 10−16