GATE CS 2020 Q.39 (CS Q.29): Number Representations and Computer Arithmetic

Consider three registers R1, R2, and R3 that store numbers in IEEE-754 single precision floating point format. Assume that R1 and R2 contain the values (in hexadecimal notation) 0x42200000 and 0xC1200000, respectively.

If $\text{R3} = \frac{\text{R1}}{\text{R2}}$ , what is the value stored in R3?

0x40800000

0xC0800000

0x83400000

0xC8500000

Official answer and solution

Official answer: B
From: Official GATE 2020 answer key

Solution

Decode the IEEE-754 single precision floating point numbers, perform the division, and re-encode the result.

Decode R1 (0x42200000):

In binary: 0 10000100 01000000000000000000000
Sign bit: $s = 0$ (positive)
Biased exponent: $E = 10000100_2 = 132_{10} \implies$ Unbiased exponent $e = 132 - 127 = 5$
Mantissa: $M = 1.01_2 = 1 + 0.25 = 1.25$
Value: $\text{R1} = +1.25 \times 2^5 = 1.25 \times 32 = 40.0$

Decode R2 (0xC1200000):

In binary: 1 10000010 01000000000000000000000
Sign bit: $s = 1$ (negative)
Biased exponent: $E = 10000010_2 = 130_{10} \implies$ Unbiased exponent $e = 130 - 127 = 3$
Mantissa: $M = 1.01_2 = 1.25$
Value: $\text{R2} = -1.25 \times 2^3 = -1.25 \times 8 = -10.0$

Perform division:

\text{R3} = \frac{\text{R1}}{\text{R2}} = \frac{40.0}{-10.0} = -4.0

Encode $-4.0$ into IEEE-754 single precision format:

Sign bit: $s = 1$ (since number is negative)
Magnitude: $4.0 = 1.0_2 \times 2^2$
Biased exponent: $E = 2 + 127 = 129_{10} = 10000001_2$
Fraction field: 23 zeros ( $0000\dots 0$ )
Combined 32-bit word: 1 10000001 00000000000000000000000
Hexadecimal: 1100 0000 1000 0000 0000 0000 0000 0000 = 0xC0800000.

Final answer

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2020 Computer Science and Information Technology (CS)
Question: 39 of 65, printed as CS Q.29
Section: Computer Science and Information Technology

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