Consider the IEEE-754 single precision floating point numbers P = 0xC1000000 and Q = 0x3F5C2EF4. Which one of the following corresponds to the product of these numbers (P x Q) in single precision format? A. 0x040C2EF4 B. 0x405C2EF4 C. 0xC0DC2EF4 D. 0xC15C2EF4

GATE 2023 · Digital Logic · IEEE Representation · medium

Answer: C. 0xC0DC2EF4

1. Decode P = 0xC1000000: 0xC1 = 1100 0001, remaining bytes = 0 Binary: 1 10000010 00000000000000000000000 Sign_P = 1, Exp_P = 10000010 = 130, Mantissa_P = 0...0 (all zeros) Significand_P = 1.0 P = -1.0 x 2^(130-127) = -1.0 x 2^3 = -8.0
2. Decode Q = 0x3F5C2EF4: 0x3F = 0011 1111, 0x5C = 0101 1100, 0x2E = 0010 1110, 0xF4 = 1111 0100 Binary: 0 01111110 10111000010111011110100 Sign_Q = 0, Exp_Q = 01111110 = 126, actual exp = -1 Mantissa_Q bits = 10111000010111011110100 Significand_Q = 1.10111000010111011110100 Q = +1.10111000010111011110100 x 2^(-1) = 0.110111000010111011110100...
3. Compute product sign and exponent: Sign_res = 1 XOR 0 = 1 (negative result) Exp_res = 130 + 126 - 127 = 129 = 10000001 in binary
4. Compute product significand: Significand_P = 1.0 (exactly) Significand_Q = 1.10111000010111011110100 Product = 1.0 x 1.10111000010111011110100 = 1.10111000010111011110100 No normalization needed (result is already in 1.xxx form) Mantissa bits (23 bits) = 10111000010111011110100
5. Assemble the result: Sign = 1, Exp = 10000001, Mantissa = 10111000010111011110100 Full 32-bit: 1 10000001 10111000010111011110100 Grouping into nibbles: 1100 0000 1101 1100 0010 1110 1111 0100 C 0 D C 2 E F 4 Result = 0xC0DC2EF4