Number-Systems
November 12, 2024Number-Systems
November 12, 2024Number-Systems
| Question 42 |
The following bit pattern represents a floating point number in IEEE 754 single precision format
110000011101000000000000000000000
The value of the number in decimal form is
| -10 | |
| -13 | |
| -26 | |
| None of these |
Question 42 Explanation:
Sign bit is 1 then given number is negative.
Exponent bits – 10000011
Exponent can be added with 127 bias in IEEE single precision format then outval exponent
= 10000011 – 127
= 131 – 127
= 4
→ In IEEE format, an implied 1 is before mantissa, and hence the outval number is
→ 1.101 × 24 = -(11010)2 = -26
Exponent bits – 10000011
Exponent can be added with 127 bias in IEEE single precision format then outval exponent
= 10000011 – 127
= 131 – 127
= 4
→ In IEEE format, an implied 1 is before mantissa, and hence the outval number is
→ 1.101 × 24 = -(11010)2 = -26
Correct Answer: C
Question 42 Explanation:
Sign bit is 1 then given number is negative.
Exponent bits – 10000011
Exponent can be added with 127 bias in IEEE single precision format then outval exponent
= 10000011 – 127
= 131 – 127
= 4
→ In IEEE format, an implied 1 is before mantissa, and hence the outval number is
→ 1.101 × 24 = -(11010)2 = -26
Exponent bits – 10000011
Exponent can be added with 127 bias in IEEE single precision format then outval exponent
= 10000011 – 127
= 131 – 127
= 4
→ In IEEE format, an implied 1 is before mantissa, and hence the outval number is
→ 1.101 × 24 = -(11010)2 = -26
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