[理工] [計組]-選擇題

[bernachom (Terry)]

 可能有一些作業系統的選擇...有些答案不是很確定，來請教一下


1.(?) The delayed branch is especially suitable for a deeper pipeline.

      // 有點看不懂這是在問什麼...

2.(X) For a pipelined processer with complicated address modes and instruction

      opcodes, finite state machine is the most appropriate approach to be

      used as a control unit.

      // pipelined應該是register?!

3.(X) By Amdahl's Law, if you invent a new method which can contribute 100%

      improvement on the 50% part of the entire system, your best speedup is

      2(or 200%).

      // 大約算了一下，應該是錯的..

4.(X) On a 32-bit addition, the overflow condition occurs when the 32nd adder

      generates a carry.

      // 32nd是指32個bit都進位嗎?

5.(?) Mutual exclusion can be enforced with a general semaphore whose initial

      value is geater than 1.

     //還沒唸到...順便問一下..QQ

6.(?) The working set of a process can be changed in response to actions by

      other processes.

    //感覺是錯的..不是很確定..


7.(T) For kernel memory management, slab allocation is preferred to buddy

      system only because it is faster.

   // 印像中是對的..



 就這幾題了..謝謝幫忙指正.



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◆ From: 61.224.205.119
※ 編輯: bernachom       來自: 61.224.205.119       (10/11 23:19)
推 gensim:1=>false:deeper pipeline 造成更多slot
→ bernachom:看懂了...一時想不到deeper pipeline是什麼..Orz..tks..
推 gensim:5=>false:一般semaphorem要達到mutual exclusion是設1
→ yesa315:2. 應該是要用微程式 3.再怎麼增加 不可能讓改進的部分=0
→ yesa315:7. 應該是對的 buddy需要回收 而slab只要mark free就好
→ bernachom:謝謝^^
推 gensim:The slab allocation algorithm has as its principal
→ gensim:benefit that memory gets allocated in exactly the same
→ gensim: size as requested, thus no internal memory
→ gensim: fragmentation exists.