windlx

说明：  经统计，某机器14条指令的使用频度分别为：0.01,0.15,0.12,0.03,0.02,0.04,0.02,0.04,0.01,0.13,0.15,0.14,0.11,0.03。分别求出用等长码、Huffman码、只有两种码长的扩展操作码3种编码方式的操作码平均码长。 解： 等长操作码的平均码长=4位 Huffman编码的平均码长=3.38位 只有两种码长的扩展操作码的平均码长=3.4位。 9．若某机要求：三地址指令4条，单地址指令255条，零地址指令16条。设指令字长为12位．每个 地址码长为3位。问能否以扩展操作码为其编码?如果其中单地址指令为254条呢?说明其理由。 答：①不能用扩展码为其编码。 ∵指令字长12位，每个地址码占3位； ∴三地址指令最多是2^(12-3-3-3)=8条， 现三地址指令需4条, ∴可有4条编码作为扩展码， ∴单地址指令最多为4×2^3×2^3=2^8=256条， 现要求单地址指令255条，∴可有一条编码作扩展码 ∴零地址指令最多为1×2^3＝8条 不满足题目要求 ∴不可能以扩展码为其编码。 ②若单地址指令254条，可以用扩展码为其编码。 ∵依据①中推导，单地址指令中可用2条编码作为扩展码 ∴零地址指令为2×2^3＝16条，满足题目要求
(By the statistics, a machine 14 commands the use of frequency were: 0.01,0.15,0.12,0.03,0.02,0.04,0.02,0.04,0.01,0.13,0.15,0.14,0.11,0.03. Were obtained using such a long code, Huffman code, there are only two yards of the three kinds of expansion opcode opcode encoding of the average code length)