作者bernachom (Terry)
看板Grad-ProbAsk
标题[理工] [计组]-选择题
时间Sun Oct 11 23:19:17 2009
可能有一些作业系统的选择...有些答案不是很确定，来请教一下
1.(?) The delayed branch is especially suitable for a deeper pipeline.
// 有点看不懂这是在问什麽...
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应该是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%).
// 大约算了一下，应该是错的..
4.(X) On a 32-bit addition, the overflow condition occurs when the 32nd adder
generates a carry.
// 32nd是指32个bit都进位吗?
5.(?) Mutual exclusion can be enforced with a general semaphore whose initial
value is geater than 1.
//还没念到...顺便问一下..QQ
6.(?) The working set of a process can be changed in response to actions by
other processes.
//感觉是错的..不是很确定..
7.(T) For kernel memory management, slab allocation is preferred to buddy
system only because it is faster.
// 印像中是对的..
就这几题了..谢谢帮忙指正.
--
※ 发信站: 批踢踢实业坊(ptt.cc)
◆ From: 61.224.205.119
※ 编辑: bernachom 来自: 61.224.205.119 (10/11 23:19)
1F：推 gensim:1=>false:deeper pipeline 造成更多slot 10/12 00:05
2F：→ bernachom:看懂了...一时想不到deeper pipeline是什麽..Orz..tks.. 10/12 00:09
3F：推 gensim:5=>false:一般semaphorem要达到mutual exclusion是设1 10/12 00:12
4F：→ yesa315:2. 应该是要用微程式 3.再怎麽增加 不可能让改进的部分=0 10/12 11:07
5F：→ yesa315:7. 应该是对的 buddy需要回收 而slab只要mark free就好 10/12 11:08
6F：→ bernachom:谢谢^^ 10/12 21:20
7F：推 gensim:The slab allocation algorithm has as its principal 10/12 23:14
8F：→ gensim:benefit that memory gets allocated in exactly the same 10/12 23:15
9F：→ gensim: size as requested, thus no internal memory 10/12 23:16
10F：→ gensim: fragmentation exists. 10/12 23:16