GATE CS 2026 Set 2 — Question 44
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Computer Organization & Architecture → Instructions & Addressing Modes → Instruction Formats
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Question
Consider a processor that has 16 general purpose registers and it uses 2-byte instruction format for all its instructions. Variable-sized opcodes are permitted. There are three different types of instructions; M-type, R-type, and C-type. Each M-type instruction has 2 register operands and a 6-bit immediate operand. Each R-type instruction has 3 register operands. Each C-type instruction has a register operand and a 6-bit offset value. If there are 2 unique M-type opcodes and 7 unique R-type opcodes, which one of the following options gives the maximum number of unique opcodes possible for C-type instructions?
Correct answer
(B) 4
Solution
Total instruction length = 16 bits.
Number of registers = 16 4 bits per register operand.M-type:
Number of registers = 16 4 bits per register operand.M-type:
- Operands: 2 registers (8 bits) + 6-bit immediate = 14 bits.
- Opcode bits: bits.
- Space consumed: .
- Operands: 3 registers (12 bits).
- Opcode bits: bits.
- Space consumed: .
- Operands: 1 register (4 bits) + 6-bit offset = 10 bits.
- Opcode bits: bits.
- Let be the number of C-type opcodes.
- Space consumed: .
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