[試題] 105上 鄭卜壬 系統程式設計 期中考

[xavier13540 (柊 四千)]

課程名稱︰系統程式設計
課程性質︰資訊工程學系大二必修
課程教師︰鄭卜壬
開課學院：電機資訊學院
開課系所︰資訊工程學系
考試日期（年月日）︰2016/11/16
考試時限（分鐘）：180

試題 :


1. (17 pts) Explanation (as clearly as possible for full credits).

   (a) (4 pts) System calls allow processes to trap into the kernel. Give two
       examples of system calls to explain under what circumstances one would
       trigger a context switch and the other would not.

   (b) (3 pts) Explain under what circumstances advisory lock is safe even
       though there are other processes probably violating a lock and trying to
       access the locked file.

   (c) (3 pts) Explain why communication through a pipe should be limited to
       processes that descend from a common ancestor.

   (d) (3 pts) Exaplain under what circumstances a UNIX-like system would
       automatically turn off the set-group-ID bit of a newly created file.

   (e) (4 pts) Explain why it's common for a process to create and open a new
       file and then immediately unlink it; however, it's not very useful for a
       process to create and open a directory and then immediately remove it.


2. (20 pts) The cp utility copies the content of a source file to a target file.
   When cp implemented with unbuffered I/O system calls, the following four
   factors (A)~(D) would significantly affect its execution time. Please answer
   the questions.

       (A) The number of the while loops
           -- Each loop copies partial file content with read() or write()
       (B) The time to wait for data ready in memory
       (C) The time to copy data from kernel's buffer cache to user's buffer and
           vice versa
           -- User's buffer denotes the buffer specified in read() and write()
       (D) The time to move data from kernel's buffer cache to disk and vice
           versa

   (a) (2 pts) What factor(s) will significantly affect user CPU time?

   (b) (2 pts) What factor(s) will significantly affect system CPU time?

   (c) (2 pts) What factor(s) will significantly affect clock/response time?

   (d) (2 pts) What factor(s) will be significantly affected when nonblocking
       I/O is taken into account, compared with blocking I/O?

   (e) (2 pts) What time (user CPU time, system CPU time, and clock/response
       time) will be significantly affected by system call fsync()?

   (f) (6 pts) Suppose the target file is redirected to null device /dev/null.
       We run the cp utility with different buffer sizes in blocking mode and
       get the following results.

    ────────────────────────────────────
    Buffer size     User CPU (sec)      System CPU (sec)    Clock Time (sec)
    ────────────────────────────────────
          1            124.89              161.65              288.64
          2             63.10               80.96              145.81
         32              4.13                5.01                9.76
        128              1.01                1.27                6.82
       4096              0.03                0.16                6.86
       8192              0.01                0.18                6.67
    ────────────────────────────────────

       (f.1) Increasing the buffer size beyond 4096 has little positive effect
             on system CPU time. Why?

       (f.2) When the buffer size is small, the difference between clock time
             and total CPU time (user CPU + system CPU) is also small, e.g., the
             sum of 124.89 and 161.65 is close to 288.64. But such difference
             increases significantly when the buffer size comes to 128. Why?

   (g) (4 pts) Suppose the time to read the source file is ignored. We run the
       cp utility with different sizes of source files in blocking mode and get
       the following results.

    ────────────────────────────────────
    Source size     User CPU (sec)    System CPU (sec)      Clock Time (sec)
    ────────────────────────────────────
       10M              0.000             0.008                 0.010
      300M              0.000             0.236                 0.239
     1000M              0.000             0.824                 0.827
    ────────────────────────────────────

       Disk I/O is assumed to be time comsuming; however, clock time is close to
       system CPU time for different sizes. Why?


3. (16 pts) Alice and Bob propose a method to share files securely. Each of them
   creates two directories OutBox and InBox in her/his home directory. OutBox is
   the place to store the files to be shared; InBox is the place to store the
   links pointing to the shared files. If Alice wants to share her file hw with
   Bob, she can just put the file in ~Alice/OutBox and then create a link
   pointing to ~/Alice/OutBox/hw in ~Bob/InBox. For security issue, no one is
   able to traverse both of OutBox and InBox except the owner. Only the owner
   can remove his/her own files from OutBox. Assume Alice and Bob belong to
   different groups and they both have no superuser privileges. Please answer
   the following questions.

   (a) (2 pts) What is the advantage of creating a link in InBox over copying a
       file to InBox?

   (b) (3 pts) Alice can put a symbolic link or a hard link in Bob's InBox.
       Which choice is better? Explain your answer.

   (c) (6 pts) What minimum access rights (i.e., read, write and execute) should
       be used for the following directories? Only consider the others class.
       Your answer should be in the form of "rwx", "r-x", or the like.

       (1) Directory InBox

       (2) Directory OutBox if hard links are adopted

       (3) Directory OutBox if symbolic links are adopted

   (d) (2 pts) If a file put in Bob's InBox is owned by Alice and its permission
       is "r-- --- ---," can Bob remove the file from his InBox directory? Why?

   (e) (3 pts) One major weakness of the method is that: (1) anyone may create
       files in ~Bob/InBox, and (2) the owner of the link put in ~Bob/InBox may
       not be Bob, who feels uncomfortable with this. Please develop a set-user-
       id program to help them solve these problems. No code is required here.
       Just describe how your method works in detail.


4. (10 pts) Alice plans to develop a function, which opens a unique temporary
   file in read/write mode and then returns its file descriptor. In her program,
   as shown below, tmpnam() is invoked to obtain a pointer pointing to a valid
   filename that doesn't exist at the time it is called. Error checking is
   ignored.

    int temp_file() {
        char *fnptr;
        int fd;
        fnptr = tmpnam (NULL);
        fd = open (fnptr, O_CREAT | O_TRUNC | O_RDWR, 0600);
        return fd;
    }

   (a) (3 pts) Specify race condition.

   (b) (3 pts) Give an example to explain why the code leads to a race
       condition.

   (c) (4 pts) Propose a method to fix the problem without changing the way to
       call tmpnam() and open() in the code. That is, you cannot replace the
       arguments of the two functions with others. What system calls will be
       called in your method? Describe how your method works in detail.


5. (19 pts) Double fork is a common solution to the problem: the parent doesn't
   need to wait for the child to complete and the child doesn't become a zombie.

   (a) (4 pts) Give tow reasons to explain why UNIX-like systems need zombie
       processes.

   (b) (3 pts) How does the copy-on-write technique improve the efficiency of
       fork()?

   (c) (12 pts) Please implement function double_fork() with the prototype:

        pid_t double_fork()

       Your code should follow the rules:
        (1) Call both fork() and vfork(). Each is invoked exactly once. Do not
            call fork() twice.
        (2) Return 0 in the child. Return the process ID of the grandchild in
            the parent.
        (3) Use a pipe and unbuffered I/O in blocking mode to avoid a race
            condition.
        (4) Cannot produce any zombie processes.
        (5) Close all of the unused file descriptors.
       The errors returned from system calls can be ignored.


6. (18 pts) Alice and Bob are writing two programs to play chess in UNIX
   environment. In this game, their programs must alternately move one piece at
   a time until one is checkmated. Assume two functions, thinking() and making_
   a_move(), are provided already. The function thinking() is to evaluate how
   good further moves would be; making_a_move() is just to move one piece. After
   calling making_a_move(), one should inform the other "it's your turn now."
   One simple solution to process synchronization is using pipe, as shown below,
   where process Alice calls write() to send message 'P' to Bob through standard
   output. Bob will be woken up by reading the message from standard input and
   make the next move.

    void main(void)
    {   // Program Alice
        char buf;
        while (1) {
            thinking();
            making_a_move();
            buf='P';
            write(STDOUT_FILENO, &buf, 1);
            read(STDIN_FILENO, &buf, 1);
        }
    }

    void main(void)
    {   // Program Bob
        char buf;
        while (1) {
            read(STDIN_FILENO, &buf, 1);
            thinking();
            making_a_move();
            buf='P';
            write(STDOUT_FILENO, &buf, 1);
        }
    }

   Please use pipe(), fork(), dup2(), execv(), and wait() to write a program,
   which forks two processes that can talk to each other with pipes. One process
   execute program Alice; the other executes program Bob. Your code doesn't need
   completeness and can ignore error returns. Unused file descriptors should be
   closed. Zombie processes are not allowed.


// 題目後面有附上幾個函式的prototype 這邊略去

--
第01話 似乎在課堂上聽過的樣子           第02話 那真是太令人絕望了
第03話 已經沒什麼好期望了               第04話 被當、21都是存在的
第05話 怎麼可能會all pass               第06話 這考卷絕對有問題啊
第07話 你能面對真正的分數嗎             第08話 我，真是個笨蛋
第09話 這樣成績，教授絕不會讓我過的     第10話 再也不依靠考古題
第11話 最後留下的補考                   第12話 我最愛的學分

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※ dyadi:轉錄至看板 b05902xxx                                      11/08 16:40