GATE CS 2026 Set 2 Q34 — Machine Instructions and Addressing Modes

To find the maximum number of unique opcodes for C-type instructions, we use the expanding opcode technique.

1. Basic Parameters:

• Instruction size = $2\text{ bytes} = 16\text{ bits}$.
• Number of registers = $16$, so each register operand requires $\log_2(16) = 4\text{ bits}$.

2. Opcode Sizes for Each Type:

• M-type: Requires $2$ registers ($2 \times 4 = 8\text{ bits}$) and a $6$-bit immediate.
  $\text{Opcode size} = 16 - (8 + 6) = 2\text{ bits}$.
  Total possible 2-bit opcodes = $2^2 = 4$. Since $2$ are used, remaining unused 2-bit prefixes = $4 - 2 = 2$.
• R-type: Requires $3$ registers ($3 \times 4 = 12\text{ bits}$).
  $\text{Opcode size} = 16 - 12 = 4\text{ bits}$.
  Using the $2$ remaining 2-bit prefixes, total possible 4-bit opcodes = $2 \times 2^{4-2} = 8$.
  Since $7$ are used, remaining unused 4-bit prefixes = $8 - 7 = 1$.
• C-type: Requires $1$ register ($4\text{ bits}$) and a $6$-bit offset.
  $\text{Opcode size} = 16 - (4 + 6) = 6\text{ bits}$.
  Using the $1$ remaining 4-bit prefix, maximum possible 6-bit opcodes = $1 \times 2^{6-4} = 2^2 = 4$.

Thus, the maximum number of unique opcodes possible for C-type instructions is 4.