Floating-Point Examples (Cont.)
Example IV
Show the IEEE 754 binary representation of the number -6.8125 in single and double precision.
Answer:
Fraction bits can be obtained by repeatedly multiplying each fraction by 2 and collecting the integer until the fractional part is 0:
0.8125 × 2 = 1.625 ⇒ 1
0.6250 × 2 = 1.250 ⇒ 1
0.2500 × 2 = 0.500 ⇒ 0
0.5000 × 2 = 1.000 ⇒ 1
So the binary strings of the fraction and number are as follows:
0.8125 = 13/16 = ½ + ¼ + 1/16 = 0.11012
-6.8125 = -110.11012 = -1.1011012×22
The exponent (single precision) = 2 + Bias = 2 + 127 = 129 and
The exponent (double precision) = 2 + Bias = 2 + 1023 = 1025.
Therefore, the single-precision representation of -6.8125 is as follows:
| 1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
and the double-precision representation of -6.8125 is as follows:
| 1 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
1 |
1 |
0 |
1 |
1 |
0 |
1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
| 0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |