GATE CS 2020 Q29 — Number Representations (Fixed and Floating Point)

To find the value stored in $R3$, we first decode the IEEE-754 single-precision floating-point values in $R1$ and $R2$.

1. Decoding $R1 = \text{0x42200000}$:

• Binary representation: $0100\ 0010\ 0010\ 0000\ 0000\ 0000\ 0000\ 0000_2$
• Sign bit ($S$) = $0$ (positive)
• Biased Exponent ($E$) = $1000\ 0100_2 = 132_{10}$
• Actual Exponent ($e$) = $132 - 127 = 5$
• Mantissa ($M$) = $1.01_2 = 1.25_{10}$
• Value of $R1 = +1.25 \times 2^5 = 40$

2. Decoding $R2 = \text{0xC1200000}$:

• Binary representation: $1100\ 0001\ 0010\ 0000\ 0000\ 0000\ 0000\ 0000_2$
• Sign bit ($S$) = $1$ (negative)
• Biased Exponent ($E$) = $1000\ 0010_2 = 130_{10}$
• Actual Exponent ($e$) = $130 - 127 = 3$
• Mantissa ($M$) = $1.01_2 = 1.25_{10}$
• Value of $R2 = -1.25 \times 2^3 = -10$

3. Calculating $R3 = \frac{R1}{R2}$:

$$R3 = \frac{40}{-10} = -4$$

4. Encoding $-4$ into IEEE-754 format:

• $-4 = -1.0 \times 2^2$
• Sign bit ($S$) = $1$
• Actual Exponent ($e$) = $2 \implies$ Biased Exponent ($E$) = $2 + 127 = 129_{10} = 1000\ 0001_2$
• Fraction ($F$) = $000\dots0_2$
• Combined Binary: $1100\ 0000\ 1000\ 0000\ 0000\ 0000\ 0000\ 0000_2$
• Hexadecimal representation: $\text{0xC0800000}$