Consider a floating-point representation in which the exponent is stored in…
1996
Consider a floating-point representation in which the exponent is stored in 2's complement form and the mantissa is stored in sign-magnitude form with 23 magnitude bits. What is the range of the magnitude of the normalized mantissa in this representation?
- A.
0 to 1
- B.
0.5 to 1
- C.
2⁻²³ to 0.5
- D.
0.5 to (1 − 2⁻²³)
Attempted by 16 students.
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Correct answer: D
In sign-magnitude mantissa representation, the sign is handled separately, so the magnitude is determined by the magnitude bits.
For a normalized binary mantissa, the first magnitude bit after the binary point must be 1. Therefore, the smallest normalized magnitude is:
0.100...₂ = 1/2 = 0.5.
With 23 magnitude bits, the largest normalized mantissa is obtained when all 23 magnitude bits are 1:
0.111...111₂ = 2⁻¹ + 2⁻² + ... + 2⁻²³ = 1 − 2⁻²³.
The exponent representation affects the scaling of the final floating-point value, but it does not change the allowed normalized range of the mantissa magnitude. Hence the range is 0.5 to (1 − 2⁻²³).