Consider the IEEE-754 single precision floating point numbers P=0xC1800000 and…
2023
Consider the IEEE-754 single precision floating point numbers
P=0xC1800000 and Q=0x3F5C2EF4.
Which one of the following corresponds to the product of these numbers (i.e., P × Q), represented in the IEEE-754 single precision format?
- A.
0x404C2EF4
- B.
0x405C2EF4
- C.
0xC15C2EF4
- D.
0xC14C2EF4
Attempted by 107 students.
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Correct answer: C
Concept: In IEEE-754, a value is sign × 1.fraction × 2(E−bias). Multiplying a float by an exact power of two does not touch the fraction bits at all: it only flips the sign for a negative factor and shifts the unbiased exponent by the power. So scaling Q by −16 = −24 means flip the sign, add 4 to Q's exponent, and copy Q's fraction unchanged.
Decode P = 0xC1800000: sign = 1 (negative); exponent field 0x83 = 131, so E = 131 − 127 = 4; fraction = 0. Hence P = −1.0 × 24 = −16, an exact power of two.
Decode Q = 0x3F5C2EF4: sign = 0 (positive); exponent field 0x7E = 126, so E = 126 − 127 = −1; fraction field = 0x5C2EF4. Hence Q = +1.fraction × 2−1 ≈ 0.86.
Apply the rule: P × Q = −24 × Q. New sign = 1 (negative). New exponent E′ = −1 + 4 = 3. Fraction stays 0x5C2EF4. No rounding is needed because a power-of-two scale never disturbs the mantissa.
Reassemble the 32-bit word: sign bit = 1; biased exponent field = E′ + 127 = 3 + 127 = 130 = 0x82; fraction = 0x5C2EF4. Packing sign·exponent·fraction gives 0xC15C2EF4.
Cross-check: Decimally, −16 × 0.86009… = −13.7615…, and decoding 0xC15C2EF4 back yields exactly −13.7615…, confirming the result.