Question

7. On a machine, an instruction is always 16 bits long, and the re are 16 registers. (1) Briefly explain the idea of expanding opcode. (5%) (2) Given this scheme of expanding opcode, identify how many three-address instructions, two-address instructions, one-address instructions, and zero- address instructions are supported. (10%) O 0000 xxxx yyyy zzzz 0001 xxxX yyyy ZZZZ 0010 xxxx yyyy zzZZZ 92 3-address SE instructions 0111 xxxx yyyy zzzz 1000 0000 yyyy zzZz 1000 0001 yyyy zzzz 1000 0010 yyyy zzzZ 2-address instructions 1111 1101 yyyy zzzz 1111 1110 0000 zzzz 1111 1110 0001 zzzz 1111 1110 0010 zzzz 1-address instructions 1111 1111 1110 zzzz

Answer 1

Consider an instruction length is 16 bits and there are 16 registers. (1) Expanding opcode: For the 16 bits long instruction, the opcode is 4 bits and the addresses of the registers are 4- bits long as there are 16 registers. Therefore, the instruction has 4-bit opcode and three 4-bit addresses The 16 three-address instruction format is shown below: 15 14 13 12 11 10 9 7 5 43 2 1 0 8 6 Address 3 Opcode Address 2 Address 1 The opcode needs to expand for 15 three-address instructions For opcode 15, the bits 8 to 15 indicates the opcode instead of 12 to 15. The bits 0 to 3 and 4 to 7 indicate two addresses

(2) For, three-address instructions: Out of 16 bits in the instruction, 12 bits is used to indicate three addresses of the registers. Each address is of 4 bits. The remaining four bits indicate opcode of the instruction. The opcode value of the first instruction is 0000 (0 in decimal) and the last instruction is 0111 (7 in decimal) The number of three-address instructions is given as 7 - 0 +1 8 Therefore, the number of three-address instructions is 8 instructions For, two-address instructions: Out of 16 bits in the instruction, 8 bits is used to indicate two addresses of the registers. Each address is of 4 bits. The remaining 8 bits indicate opcode of the instruction The opcode value of the first instruction is 10000000 (128 in decimal) and the last instruction is 11111101 (253 in decimal) The number of two-address instructions is given as 253 128 +1 126 Therefore, the number of two-address instructions is 126 instructions.

For, one-address instructions: Out of 16 bits in the instruction, 4 bits is used to indicate one address of a register. The remaining 12 bits indicate opcode of the instruction. The opcode value of the first instruction is 111111100000 (4064 in decimal) and the last instruction is 11111111 11 10(4094 in decimal) The number of one-address instructions is given as 4094 4064 +1 126 Therefore, the number of one-address instructions is 31 instructions. For, zero-address instructions: Out of 16 bits in the instruction, the 16 bits indicate opcode of the instruction The opcode value of the first instruction is 11111111111 10000 (65520 in decimal) and the last instruction is 111111 11 1 1 1 1 1 1 1 1 (65535 in decimal) The number of zero-address instructions is given as 65535- 65520 +1 126 Therefore, the number of zero-address instructions is 16 instructions