作者查詢 / aiweisen
作者 aiweisen 在 PTT [ Grad-ProbAsk ] 看板的留言(推文), 共42則
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1F推:妳先想看看context switch03/25 11:43
35F推:看完這篇 我中山沒希望了= =02/28 21:36
2F推:我離散一看到考卷就腿軟了 沒一題會寫= =02/26 21:24
3F推:能用MASTER就用吧 怕不夠嚴謹就先寫定義 在by master02/21 19:46
17F推:第三題好像是圖論的02/13 22:47
1F推:我很久沒寫程式了 所以不知道有沒有想錯 我是認為02/11 12:32
2F→:Test_and_Set(lock)不是VOID 是有傳回值 所以需要一個變02/11 12:33
3F→:數來接收Test_and_Set(lock)的結果 當然,如果我的想法02/11 12:35
4F→:錯誤,那麼再從另一個角度來看。第二個while如果卡住了02/11 12:36
5F→:要它執行什麼事情?應該要插NOOP 不過這裡我不知道能不能02/11 12:36
6F→:= = 所以我想講義只用第一個While不用第二個 可能是為了02/11 12:37
7F→:增加可讀性而已 如果拿第二個跟我同學解釋 我也不太會解02/11 12:39
8F→:釋....囧02/11 12:39
8F→:如同一樓說的 是sum delay or carry delay的問題01/08 22:27
9F→:4-bit ripple-carry adder = 8倍延遲 是因為CarryOut是201/08 22:28
10F→:個 gate dely,又因為ripple-carry的特色是串連,所以必01/08 22:30
11F→:須經過8個gate dely Cout才會對(因為是4個加法器)01/08 22:32
12F→:而你提到CLA只花到6倍延遲,那個指的是sum dely。因為只01/08 22:36
13F→:花一個dely產生了gi跟pi(這裡加法器的and/or gate是平行01/08 22:38
14F→:處理;gi=ai‧bi,pi=ai+bi)而再花2個dely產生CarryOut。所01/08 22:40
15F→:以要產生完整的Cout要花3個dely(例如 c1=g0+p0‧c0)01/08 22:42
16F→:而sum dely要產生,除了要看原本自己的gate dely外,還要01/08 22:44
17F→:考慮Cout的gate dely。sum 的gate dely是3,再加上Cout所01/08 22:46
18F→:以是3+3=6個gate dely。另外前面的ripplecarry的sum dely01/08 22:47
19F→:是9個gate dely(因為輸出s3,要先保證前面c1c2c3都正確,01/08 22:49
20F→:s3才會正確,所以是2+2+2+3=9)01/08 22:49
27F推:當初我上張凡他是這麼說的 產生qi與pi只花一個dely完成。01/08 22:59
28F→:那個Add 裡面只有and與or這兩個gate,而且是各做各的沒有01/08 23:00
29F→:相互等待的問題。就像我上面提到gi=ai‧bi pi=ai+bi01/08 23:01
39F推:感謝holder指正,其實老師也沒說這麼詳細,我會去翻你說01/08 23:15
40F→:的這兩個差異 感謝01/08 23:15
41F推:阿...沒有定義sum delay or carry delay的話,可能要請01/08 23:18
43F→:holder319大幫忙補充了...冏"01/08 23:18
1F推:2^10算08/27 18:23
2F→:抱歉 因為我沒買過第三版的 所以不知道那是第三版的08/16 16:59
3F→:如果都是課本的 那就沒問題了08/16 17:00
1F推:硬碟是I/O,memory是為了增加存取速度用的,否則直接存取08/10 23:55
2F→:會很慢,不符合效益。08/10 23:56
3F→:順道一題,兩者速度不一樣,memory比disk快幾百萬倍08/10 23:58