推 skyHuan: 16 instrs. => op code要4碼 04/20 12:14
→ skyHuan: 32 regs => rs/rt要5碼 04/20 12:14
→ skyHuan: 所以此type的指令需要4+5+5=14碼 04/20 12:14
今天看到一題如下，請問要如何計算？
If the computer has 32 data registers (R0 to R31), 2000 words in memory,
and 16 different instructions. What is the minimum size of an add instruction
in bits if a typical instruction use the following format: add R0 R2
(A) 14
(B) 13
(C) 28
(D) 29
答案是給A 請問是如何得出的？
感謝各位！
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