update grades and todos.

Signed-off-by: Certseeds <51754303+Certseeds@users.noreply.github.com>
Certseeds · Jul 4, 2020 · ad3a8fd · ad3a8fd
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diff --git a/lab05_report03/src/scheduler.c b/lab05_report03/src/scheduler.c
@@ -9,6 +9,8 @@
 #include <fcntl.h>
 #include <time.h>
 #include "job.h"
+// TODO, only get 0 of 20 points.
+// if you know why,plase create issues.
 void pid_to_name(int pid, char *ret);
 void pid_to_name_2(int pid, char *ret);
 

diff --git a/lab06_report04/src/init.h b/lab06_report04/src/init.h
@@ -4,7 +4,7 @@
  * @Author: nanoseeds
  * @Date: 2020-04-07 14:54:54
  * @LastEditors: nanoseeds
- * @LastEditTime: 2020-04-07 20:17:46
+ * @LastEditTime: 2020-07-04 17:57:46
  */
 #include <fcntl.h>
 #include <pthread.h>
@@ -13,6 +13,6 @@
 #include <stdlib.h>
 #include <sys/stat.h>
 #include <unistd.h>
-
+// 拿到了50/50.
 extern sem_t db, rc;
 extern int readcount;
diff --git a/lab08_report05/src/banker.cpp b/lab08_report05/src/banker.cpp
@@ -21,6 +21,7 @@
  * @Date: 2020-04-13 19:19:15 
  * @LastEditors : nanoseeds
  */
+// TODO, 只拿到了40分中的20分.
 #include <iostream>
 #include <string>
 #include <vector>

diff --git a/lab09_report06/report06.md b/lab09_report06/report06.md
@@ -4,7 +4,7 @@
  * @Author: nanoseeds
  * @Date: 2020-04-23 10:26:20
  * @LastEditors: nanoseeds
- * @LastEditTime: 2020-06-04 22:51:05
+ * @LastEditTime: 2020-07-04 17:54:08
  * @License: CC-BY-NC-SA_V4_0 or any later version 
  -->
 0. 
@@ -20,7 +20,7 @@
 
 3. What is the segmentation mechanism and its advantages & disadvantages:
  1. 分段机制将内存分成长度不定的段,并将每一段中的虚拟地址空间映射到物理地址空间(每一段中对应的物理地址是连续的).
- 2. 分段机制的优点:将地址空间隔离,进程之间不会互相影响;运行时地址较为确定,方便使用.
+ 2. 分段机制的优点:将地址空间隔离,进程之间不会互相影响;运行时地址较为确定,方便使用;可以方便的与其他用户共享代码和数据.
  3. 缺点:由于段的长度不确定,所以会产生碎片.
 
 4. What is the paging mechanism and its advantages & disadvantages:

diff --git a/lab09_report06/src/mm.cpp b/lab09_report06/src/mm.cpp
@@ -18,7 +18,7 @@
  * 实现了相邻free_block的merge.
  * 每次打印出的free_block均为排序后的顺序.(BF为size小到大,WF为size大到小,FF为地址低位->高位)
  * */
-
+// 完成了基本功能,60分拿到了50.
 
 using std::cout;
 using std::cin;

diff --git a/report_07/report_07.md b/report_07/report_07.md
@@ -4,9 +4,10 @@
  * @Author: nanoseeds
  * @Date: 2020-05-10 18:42:40
  * @LastEditors: nanoseeds
- * @LastEditTime: 2020-06-04 22:51:20
+ * @LastEditTime: 2020-07-04 17:48:25
  * @License: CC-BY-NC-SA_V4_0 or any later version 
  -->
+
 1. FIFO 
 FIFO算法,每次先查看是否在内存中,如果在内存中则不变;
 若不在内存中,如果队列已满,就踢走队列头部的页面,将页面调入队尾
@@ -31,8 +32,10 @@ FIFO算法,每次先查看是否在内存中,如果在内存中则不变;
 直到发现页为0,置换此页.节约了在两个列表之间转换的成本.
 但是每次还是需要遍历,需要O(n)的时间复杂度.
 
-1. second-chance算法: 
+5. second-chance算法: 
 为了避免FIFO把经常使用的页面踢出的弊端,使用一个检查位,
 给经常使用的,检查位为1的页面重新加入列表,给第二次机会.从而提升了效率.
 换言之,使用了FIFO模式和LRU模式相结合.
-从实现上来看,每次都要遍历需要O(n)的时间复杂度.
+从实现上来看,每次都要遍历需要O(n)的时间复杂度.
+
+PS:由于不知名原因,这个report没有被评分,不知道具体情况.
diff --git a/report_07/src/pr.cpp b/report_07/src/pr.cpp
@@ -80,7 +80,8 @@ static std::unordered_map<Algorithm, std::function<void()>> umap = {
  {MIN, MIN_F},
  {CLOCK, CLOCK_F},
  {SECOND_CHANCE, SECOND_CHANCE_F}};
-
+// TODO, this do not get any scores, i do not know why.
+// If you know, please create issues.
 int main() {
  read_input();
  umap[algorithm]();

diff --git a/report_08/report_08.md b/report_08/report_08.md
@@ -4,7 +4,7 @@
  * @Author: nanoseeds
  * @Date: 2020-05-19 17:14:20
  * @LastEditors: nanoseeds
- * @LastEditTime: 2020-05-24 21:25:24
+ * @LastEditTime: 2020-07-04 17:41:00
  * @License: CC-BY-NC-SA_V4_0 or any later version 
  -->
 [Toc]
@@ -88,4 +88,5 @@ Answer:
  2. 总共需要的时间为寻道时间+旋转等待时间,旋转等待时间中,每分钟12000转,平均每秒200转,每转平均5ms.而需要进行的旋转有五次,所以一共30ms.如果将199到0的过程认为消耗寻道时间的话,一共消耗(199-100)+(90-0)+(199-0)=388ms,如果不认为的话,一共小号(199-100)+(90-0) =189ms.
  所以如果不认为从最高磁道到最低磁道消耗时间需要计算的话,一共消耗219ms,如果认为需要消耗的话(实际上,这个观点来自Lab10 Disk Scheduling guide的样例,在资料中C-look就计算了这个时间,并且CLook作为C-Scan的改进,很难想象花费的时间比C-Scan还多),一共需要418ms.
  PS:另一种思路是每转平均需要2.5ms,所以是388+15=403.
+ fix: 正确答案为403ms.
 2. 如果使用SSD,当然要使用FIFO算法,因为SSD的寻道时间几乎可以忽略不计,所以这些优化方式都没有必要进行.
diff --git a/report_08/src/disk.cpp b/report_08/src/disk.cpp
@@ -167,8 +167,12 @@ void LOOK(deque<int32_t> track) {
  output("LOOK", track, distance);
 }
 
+// TODO, needed is only C_LOOK algorith.
+// BUT i only get 26/40 scores, why?
+// IF you know why, please create issue for repo.
 void C_LOOK(deque<int32_t> track) {
  // TODO , infact, distance of lowest back to highest is hard to decide to calcaute or not.
+ // this refer the example.in.txt and example.out.txt.
  std::sort(std::begin(track), std::end(track), std::greater<int32_t>());
  int32_t dis = S + 2 * track.front() - 2 * track.back();
  while (track.back() <= S) {