problem 1

子线程循环 10 次，接着主线程循环 100 次，接着又回到子线程循环 10 次，接着再回到主线程又循环 100 次，如此循环50次，试写出代码

pthread_cond_t pcond = PTHREAD_COND_INITIALIZER;

pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;

bool isMainThreadTurn = false;

#define operation() printf("1")

const int TOTAL_LOOP = 50;

const int SUB_OP_PER_LOOP = 10;

const int MAIN_OP_PER_LOOP = 100;

void* fun(void* arg) {

for (int loop = 0; loop < TOTAL_LOOP; ++loop) {

pthread_mutex_lock(&mutex);

while (isMainThreadTurn == true) {

pthread_cond_wait(&pcond, &mutex);

}

for (int op = 0; op < SUB_OP_PER_LOOP; ++op) {

operation();

}

printf("\n");

isMainThreadTurn = true;

pthread_mutex_unlock(&mutex);

pthread_cond_signal(&pcond);

}

return (void*)0;

}

int main() {

pthread_t tid[1];

pthread_create(tid, NULL, fun, NULL);

// can shorten this part, as basically two parts are almost the same

for (int loop = 0; loop < TOTAL_LOOP; ++loop) {

pthread_mutex_lock(&mutex);

while (isMainThreadTurn == false) {

pthread_cond_wait(&pcond, &mutex);

}

for (int op = 0; op < MAIN_OP_PER_LOOP; ++op) {

operation();

}

printf("\n");

isMainThreadTurn = false;

pthread_mutex_unlock(&mutex);

pthread_cond_signal(&pcond);

}

return 0;

}

problem 2

编写一个程序，开启3个线程，这3个线程的ID分别为A、B、C，每个线程将自己的ID在屏幕上打印10遍，要求输出结果必须按ABC的顺序显示；如：ABCABC….依次递推。

#define gettid() syscall(__NR_gettid)

pthread_cond_t pcond = PTHREAD_COND_INITIALIZER;

pthread_mutex_t pmutex = PTHREAD_MUTEX_INITIALIZER;

const int TOTAL_LOOP = 10;

const unsigned int THREAD_NUM = 3;

unsigned int ThreadTurnToPrint = 0;

__thread unsigned int tid;

#define operation() printf("|%u|", tid)

void* fun(void* arg) {

unsigned int myTurn = (unsigned int) arg;

tid = gettid();

for (int loop = 0; loop < TOTAL_LOOP; ++loop) {

pthread_mutex_lock(&pmutex);

while (myTurn != ThreadTurnToPrint) {

pthread_cond_wait(&pcond, &pmutex);

}

operation();

ThreadTurnToPrint = ( 1 + ThreadTurnToPrint) % 3;

pthread_mutex_unlock(&pmutex);

pthread_cond_broadcast(&pcond);

}

return (void*)0;

}

int main()

{

pthread_t tid[THREAD_NUM];

for (int i = 0; i < THREAD_NUM; ++i) {

pthread_create(&tid[i], NULL, fun, (void*)i);

}

/*in my way, I waste mainThread, as mainThread can also do part of job*/

for (int i = 0; i < THREAD_NUM; ++i) {

pthread_join(tid[i], NULL);

}

return 0;

}

problem 3

编写一个程序，程序会启动4个线程，向4个文件1.txt，2.txt，3.txt，4.txt里写入数据，每个线程只能写一个值。

线程A：只写1

线程B：只写2

线程C：只写3

线程D：只写4

4个文件1.txt，2.txt，3.txt，4.txt

程序运行起来，4个文件的写入结果如下：

A：12341234...

B：23412341...

C：34123412...

D：41234123...

pthread_cond_t pcond = PTHREAD_COND_INITIALIZER;

pthread_mutex_t pmutex = PTHREAD_MUTEX_INITIALIZER;

const int TOTAL_LOOP = 10;

const int THREAD_NUM = 4;

int fds[THREAD_NUM];

int turns[THREAD_NUM] = {1, 2, 3, 4};

#define operation(num) printf("%d ", num)

__thread int count[THREAD_NUM] = {TOTAL_LOOP, TOTAL_LOOP, TOTAL_LOOP, TOTAL_LOOP};

__thread char buf[1];

__thread int numToPrint;

void* fun(void* arg) {

numToPrint = (int) arg;

buf[0] = numToPrint + '0';

for (int loop = 0; loop < TOTAL_LOOP * THREAD_NUM;) {

pthread_mutex_lock(&pmutex);

while (turns[0] != numToPrint && turns[1] != numToPrint &&

turns[2] != numToPrint && turns[3] != numToPrint)

{

pthread_cond_wait(&pcond, &pmutex);

}

for (int idx = 0; idx < THREAD_NUM; ++idx) {

if (turns[idx] == numToPrint && count[idx]) {

while ( write(fds[idx], buf, sizeof(buf)) < 0 ) {

if (errno != EAGAIN) {

perror("write err.");

exit(0);

}

}

if ( ++turns[idx] > THREAD_NUM) {

turns[idx] = 1;

}

--count[idx];

++loop;

}

}

pthread_mutex_unlock(&pmutex);

pthread_cond_broadcast(&pcond);

}

return (void*)0;

}

int main()

{

char filename[]= "?.txt";

for (int i = 0; i < THREAD_NUM; ++i) {

filename[0] = i + '1';

fds[i] = open(filename, O_RDWR | O_CREAT | O_TRUNC, S_IRWXU);

if (fds[i] < 0) {

perror("open fail");

exit(0);

}

}

pthread_t tid[THREAD_NUM];

for (int i = 0; i < THREAD_NUM; ++i) {

pthread_create(&tid[i], NULL, fun, (void*)(i + 1));

}

for (int i = 0; i < THREAD_NUM; ++i) {

pthread_join(tid[i], NULL);

}

return 0;

}