# 2018q1 第 13 週測驗題

### 測驗 `1`

考慮以下 C 程式:

#include <pthread.h>
#include <stdio.h>
#include <string.h>
#include <sys/time.h>
#include <unistd.h>
#define MAX 5
static int pid;
static pthread_t thread[2];
static pthread_mutex_t mut;
static int number = 0, i;

static void *thread1() {
    printf("thread1 : I'm thread 1\n");
    for (i = 0; i < MAX; i++) {
        printf("thread1 : number = %d\n", number);
        pthread_mutex_lock(&mut);
        number++;
        pthread_mutex_unlock(&mut);
        sleep(2);
    }
    printf("pid : %d, thread1 : done\n", pid);
    pthread_exit(NULL);
}

static void *thread2() {
    printf("thread2 : I'm thread 2\n");
    for (i = 0; i < MAX; i++) {
        printf("thread2 : number = %d\n", number);
        pthread_mutex_lock(&mut);
        number++;
        pthread_mutex_unlock(&mut);
        sleep(3);
    }
    printf("pid : %d, thread2 : done\n", pid);
    pthread_exit(NULL);
}

static void thread_create(void) {
    int temp;
    memset(&thread, 0, sizeof(thread));
    /* Create threads */
    if ((temp = pthread_create(&thread[0], NULL, thread1, NULL)))
        printf("Thread 1 create fail\n");
    else
        printf("Thread 1 create successfully\n");
    if ((temp = pthread_create(&thread[1], NULL, thread2, NULL)))
        printf("Thread 2 create fail");
    else
        printf("Thread 2 create successfully\n");
}

static void thread_wait(void) {
    /* Wait for threads */
    if (thread[0]) {
        pthread_join(thread[0], NULL);
        printf("pid - %d Thread 1 Finished\n", pid);
    }
    if (thread[1]) {
        pthread_join(thread[1], NULL);
        printf("pid - %d Thread 2 Finished\n", pid);
    }
}

int main() {
    pthread_mutex_init(&mut, NULL);
    printf("Main thread create threads\n");
    thread_create();
    pid = fork();
    printf("pid-%d Main thread wait for threads\n", pid);
    thread_wait();
    return 0;
}

在 macOS 編譯並執行，得到以下程式輸出:

Main thread create threads
Thread 1 create successfully
thread1 : I'm thread 1
thread1 : number = 0
Thread 2 create successfully
thread2 : I'm thread 2
thread2 : number = 0
pid-90400 Main thread wait for threads
pid-0 Main thread wait for threads
pid - 0 Thread 1 Finished
pid - 0 Thread 2 Finished
thread1 : number = 2
thread2 : number = 3
thread1 : number = 4
thread2 : number = 5
pid : 90400, thread1 : done
pid - 90400 Thread 1 Finished
pid : 90400, thread2 : done
pid - 90400 Thread 2 Finished

留意第 65 行 `pid = fork();` 敘述，`fork` 系統呼叫觸發後，會「分叉」執行單元為原本的 parent process 和新建立的 child process。從以下選項中挑出最符合的描述。

==作答區==
X = ?
* `(a)` 在 macOS 和 GNU/Linux 有著完全不同的表現，上述輸出不僅 PID 數值不同，連 `number =` 的數值順序也不一樣
* `(b)` 在 child process 中，不具備 `fork` 前由 parent process 建立的 2 個執行緒 (Thread 1 和 2)，這點符合 POSIX 規範中 fork 的語意
* `(c)` 受制於 mutex `mut` 的影響，致使 child process 內沒辦法繼續執行 Thread 1 和 2

參考資料: [Mutex lock for Linux Thread Synchronization](https://www.geeksforgeeks.org/mutex-lock-for-linux-thread-synchronization/)

:::info
延伸題目:
1. 解釋程式運作，擴充測試的範圍
2. 參照 [Threads and fork(): think twice before mixing them](http://www.linuxprogrammingblog.com/threads-and-fork-think-twice-before-using-them)，添增 `pthread_atfork()` 來改寫程式碼，使得 child/parent 兩個 process 的執行緒得以交替輸出
:::