c - 在具有4个线程的程序上调用pthread_cond_signal时，同一线程将获得互斥体

Question

#include <stdio.h>
#include <stdlib.h>
#include <pthread.h>

pthread_t node[4];
pthread_mutex_t token;
pthread_cond_t cond;
int id=0;

void *func(int n)
   {
        int count = 0;
        while (count < 10){
                 pthread_mutex_lock(&token);
                while (id != n){
                        printf("Whoops not my turn, id=%d\n",n);
                        pthread_cond_wait(&cond, &token);}
                //if (id == n){
                        count += 1;
                        printf ("My turn! id= %d\n",n);
                        printf("count %d\n", count);
                        if (id == 3){
                                id = 0;}
                        else{
                                id += 1;}
                        //}else{
                        //      printf("Not my turn! id=%d\n",n);}
                        //      pthread_mutex_unlock(&token);
                        //      sleep(2);}
                        pthread_mutex_unlock(&token);
                        pthread_cond_signal(&cond);}
                        printf ("ID=%d has finished\n",n);

        return(NULL);
   }

int main()
   {
   int i;
   pthread_mutex_init(&token,NULL);
        pthread_cond_init(&cond,NULL);
   for(i=0;i<4;i++)
      pthread_create(&node[i],NULL,(void *)func,(void *)i);

   for(i=0;i<4;i++)
      pthread_join(node[i],NULL);

   pthread_mutex_destroy(&token);

   return 0;
   }

这是我的代码，它是一个使用线程的C程序。我认为这里不需要知道它的目的，但是我将举一个例子说明我遇到的问题。

说id为1(在func中由n定义)的线程获取互斥锁，并返回“我的转弯!id = 1”。当调用mutex_unlock和cond_signal时，下一个获取互斥锁的线程实际上将再次是ID为1的线程，并且它将显示“Whoops not my turn，id = 1”。只有这样，ID为2的线程才能获得互斥锁，并打印“My turn!id = 2”，但是当ID = 2的线程将获得互斥锁。这是我的程序输出:

Whoops not my turn, id=1
Whoops not my turn, id=2
Whoops not my turn, id=3
My turn! id= 0
count 1
Whoops not my turn, id=0
My turn! id= 1
count 1
Whoops not my turn, id=1
My turn! id= 2
count 1
Whoops not my turn, id=2
My turn! id= 3
count 1
Whoops not my turn, id=3
My turn! id= 0
count 2
Whoops not my turn, id=0
My turn! id= 1
count 2
Whoops not my turn, id=1
My turn! id= 2
count 2
Whoops not my turn, id=2
My turn! id= 3
count 2
Whoops not my turn, id=3
My turn! id= 0
count 3
Whoops not my turn, id=0
My turn! id= 1
count 3
Whoops not my turn, id=1
My turn! id= 2
count 3
Whoops not my turn, id=2
My turn! id= 3
count 3
Whoops not my turn, id=3
My turn! id= 0
count 4
Whoops not my turn, id=0
My turn! id= 1
count 4
Whoops not my turn, id=1
My turn! id= 2
count 4
Whoops not my turn, id=2
My turn! id= 3
count 4
Whoops not my turn, id=3
My turn! id= 0
count 5
Whoops not my turn, id=0
My turn! id= 1
count 5
Whoops not my turn, id=1
My turn! id= 2
count 5
Whoops not my turn, id=2
My turn! id= 3
count 5
Whoops not my turn, id=3
My turn! id= 0
count 6
Whoops not my turn, id=0
My turn! id= 1
count 6
Whoops not my turn, id=1
My turn! id= 2
count 6
Whoops not my turn, id=2
My turn! id= 3
count 6
Whoops not my turn, id=3
My turn! id= 0
count 7
Whoops not my turn, id=0
My turn! id= 1
count 7
Whoops not my turn, id=1
My turn! id= 2
count 7
Whoops not my turn, id=2
My turn! id= 3
count 7
Whoops not my turn, id=3
My turn! id= 0
count 8
Whoops not my turn, id=0
My turn! id= 1
count 8
Whoops not my turn, id=1
My turn! id= 2
count 8
Whoops not my turn, id=2
My turn! id= 3
count 8
Whoops not my turn, id=3
My turn! id= 0
count 9
Whoops not my turn, id=0
My turn! id= 1
count 9
Whoops not my turn, id=1
My turn! id= 2
count 9
Whoops not my turn, id=2
My turn! id= 3
count 9
Whoops not my turn, id=3
My turn! id= 0
count 10
ID=0 has finished
My turn! id= 1
count 10
ID=1 has finished
My turn! id= 2
count 10
ID=2 has finished
My turn! id= 3
count 10
ID=3 has finished

如您所见，每次成功后，线程都会打印“My turn!”，此后它将得到互斥锁，并导致“Whoops not my turn!”。我不明白为什么会在调用pthread_cond_signal时发生这种情况，在当前线程可以重新获取互斥锁之前，pthread_cond_signal应该发出信号通知另一个线程要唤醒。请帮助我找到此解决方案，因为我认为我缺少一些重要的东西。如果缺少我的解释，请随时向我询问更多信息。非常感谢您的宝贵时间!

Answer 1

@rcgldr已经对所涉及的时间给出了很好的解释。如果您想增加给另一个线程一个机会的机会，请尝试添加一个对pthread_yield的调用，这应该使调度程序有机会选择另一个线程，尽管这也不是保证。