为什么即使 sem_t 值为零，sem_wait() 函数也不会阻塞？
why dosen't sem_wait() function block even when sem_t value is zero?
我正在尝试为哲学家就餐问题（有五个哲学家）实施一个简单的解决方案，我的解决方案基于以下逻辑：
sem_t S[philosophers_number]
for each philosopher
{
while(TRUE)
{
if(current philosopher number != last philosopher)
{
thinking()
//i is number of current philosopher
sem_wait(take_chopstick(S[(i+1) % philosophers_number])) // right chopstick
sem_wait(take_chopstick(S[i])) // left chopstick
eat()
sem_post(put_chopstick(S[(i+1) % philosophers_number]))
sem_post(put_chopstick(S[i]))
}
else
{
thinking()
//i is number of current philosopher
sem_wait(take_chopstick(S[i])) // left chopstick
sem_wait(take_chopstick(S[(i+1) % philosophers_number])) // right chopstick
eat()
sem_post(put_chopstick(S[i]))
sem_post(put_chopstick(S[(i+1) % philosophers_number]))
}
}
每个哲学家首先思考不到三秒钟
然后如果右边的筷子可用哲学家就拿它，如果也有左边的筷子哲学家也会拿那个并开始吃不到三秒钟
然后哲学家会放下筷子让其他哲学家可以使用
为了避免循环等待，最后一个哲学家我会先拿左边的筷子再拿右边的筷子，重复同样的过程
下面是我根据这个逻辑实现的代码：
#include <stdio.h>
#include <string.h>
#include <pthread.h>
#include <semaphore.h>
#include <unistd.h>
#include <stdlib.h>
#define THREADS 5
sem_t chopstick[THREADS];
void thinking(int ph_num)
{
printf("philosopher %d is thinking\n", ph_num);
int t = rand() % 3;
sleep(t);// up to 3 secs thinking
}
void eat(int ph_num)
{
printf("philosopher %d is eating\n", ph_num);
int t = rand() % 3;
sleep(t);// up to 3 secs eating
}
void *philosopher(void * ph_num )
{
int num=(int)ph_num;
while(1)
{
if(num < THREADS - 1)
{
thinking(num);
//pick up right chopstick
sem_wait(&chopstick[(num + 1) % THREADS]);
//to make deadlocks absolutly happen, wait 1 sec then pickup left chopstick
sleep(1);
//pick up left chopstick
sem_wait(&chopstick[num]);
eat(num);
//put down right chopstick
sem_post(&chopstick[(num + 1) % THREADS]);
//put down left chopstick
sem_post(&chopstick[num]);
}
else // last one pick left chopstick first, instead of right one to avoid cyclic wait
{
thinking(num);
//pick up left chopstick
sem_wait(&chopstick[num]);
//to make deadlocks absolutly happen, wait 1 sec then pickup left chopstick
sleep(1);
//pick up right chopstick
sem_wait(&chopstick[(num + 1) % THREADS]);
eat(num);
//put down left chopstick
sem_post(&chopstick[num]);
//put down right chopstick
sem_post(&chopstick[(num + 1) % THREADS]);
}
}
pthread_exit((void *)num);
}
int main ()
{
for(int i = 0; i < THREADS; i++)
{
sem_init(&chopstick[i],0,1);
}
pthread_t threads[THREADS];
for(int i = 0; i < THREADS; i++)
pthread_create(&threads[i], NULL, philosopher, (void *)i);
for(int i = 0; i < THREADS; i++)
pthread_join(threads[i],NULL);
return 0;
}
但是在调试这段代码的过程中发生了一个问题，其中 chopstick[i] 在 sem_wait(&chopstick[num]) 之前 0 而不是阻塞当前线程，直到筷子可用 sem_wait() 继续, 所以一位哲学家开始吃饭时没有真正的筷子。
谁能帮我看看我的问题出在哪里？
你的实现是正确的，问题出在调试方法上。如果你使用 gdb，你将只在一个线程上停止，而线程的其余部分将继续执行，所以在你检查信号量和你进入下一行的时间之间，其他线程将推进执行并可以更改您检查过的值。
为了有效调试线程，您需要确保只有当前观察到的线程被调度，其余线程被阻塞。为此，您需要在线程停止后更改 scheduler-locking。您可以将其设置为 on 或 step，具体取决于您是希望线程完全停止，还是仅在单步操作期间停止（有关详细信息，请参阅 help set scheduler-locking）。
一旦线程被锁定，您可以使用info threads 来检查当时其余线程在做什么。可以用thread <<n>>换成第n个线程，用where查看线程栈。
这里是调度程序设置为 step 的示例。您可以看到只有一个线程在 next 命令上取得进展。
(gdb) b 37
Breakpoint 1 at 0x1388: file test003.c, line 37.
(gdb) r
Starting program: /home/jordan/Development/tmptest/a.out
[Thread debugging using libthread_db enabled]
Using host libthread_db library "/lib/x86_64-linux-gnu/libthread_db.so.1".
[New Thread 0x7ffff7d90700 (LWP 4002538)]
philosopher 0 is thinking
[New Thread 0x7ffff758f700 (LWP 4002539)]
philosopher 1 is thinking
[New Thread 0x7ffff6d8e700 (LWP 4002540)]
philosopher 2 is thinking
[2] picking 3
[New Thread 0x7ffff658d700 (LWP 4002541)]
[Switching to Thread 0x7ffff6d8e700 (LWP 4002540)]
Thread 4 "a.out" hit Breakpoint 1, philosopher (ph_num=0x2) at test003.c:37
37 sem_wait(&chopstick[(num + 1) % THREADS]);
(gdb) set scheduler-locking step
(gdb) info threads
Id Target Id Frame
1 Thread 0x7ffff7d91740 (LWP 4002534) "a.out" clone () at ../sysdeps/unix/sysv/linux/x86_64/clone.S:78
2 Thread 0x7ffff7d90700 (LWP 4002538) "a.out" 0x00007ffff7e743bf in __GI___clock_nanosleep (clock_id=clock_id@entry=0, flags=flags@entry=0,
req=req@entry=0x7ffff7d8fe60, rem=rem@entry=0x7ffff7d8fe60) at ../sysdeps/unix/sysv/linux/clock_nanosleep.c:78
3 Thread 0x7ffff758f700 (LWP 4002539) "a.out" 0x00007ffff7e743bf in __GI___clock_nanosleep (clock_id=clock_id@entry=0, flags=flags@entry=0,
req=req@entry=0x7ffff758ee60, rem=rem@entry=0x7ffff758ee60) at ../sysdeps/unix/sysv/linux/clock_nanosleep.c:78
* 4 Thread 0x7ffff6d8e700 (LWP 4002540) "a.out" philosopher (ph_num=0x2) at test003.c:37
5 Thread 0x7ffff658d700 (LWP 4002541) "a.out" clone () at ../sysdeps/unix/sysv/linux/x86_64/clone.S:78
(gdb) n
38 printf("[%i] picked %i\n", num, (num + 1) % THREADS);
(gdb) info threads
Id Target Id Frame
1 Thread 0x7ffff7d91740 (LWP 4002534) "a.out" clone () at ../sysdeps/unix/sysv/linux/x86_64/clone.S:78
2 Thread 0x7ffff7d90700 (LWP 4002538) "a.out" 0x00007ffff7e743bf in __GI___clock_nanosleep (clock_id=clock_id@entry=0, flags=flags@entry=0,
req=req@entry=0x7ffff7d8fe60, rem=rem@entry=0x7ffff7d8fe60) at ../sysdeps/unix/sysv/linux/clock_nanosleep.c:78
3 Thread 0x7ffff758f700 (LWP 4002539) "a.out" 0x00007ffff7e743bf in __GI___clock_nanosleep (clock_id=clock_id@entry=0, flags=flags@entry=0,
req=req@entry=0x7ffff758ee60, rem=rem@entry=0x7ffff758ee60) at ../sysdeps/unix/sysv/linux/clock_nanosleep.c:78
* 4 Thread 0x7ffff6d8e700 (LWP 4002540) "a.out" philosopher (ph_num=0x2) at test003.c:38
5 Thread 0x7ffff658d700 (LWP 4002541) "a.out" clone () at ../sysdeps/unix/sysv/linux/x86_64/clone.S:78
如你所见，执行next后，我一直在同一个线程，其他线程没有进行。
我已经修改了代码以使正在发生的事情更加可见，这是我使用的代码：
#include <stdio.h>
#include <string.h>
#include <pthread.h>
#include <semaphore.h>
#include <unistd.h>
#include <stdlib.h>
#define THREADS 5
sem_t chopstick[THREADS];
void thinking(int ph_num)
{
printf("philosopher %d is thinking\n", ph_num);
int t = rand() % 3;
sleep(t);// up to 3 secs thinking
}
void eat(int ph_num)
{
printf("philosopher %d is eating\n", ph_num);
int t = rand() % 3;
sleep(t);// up to 3 secs eating
}
void *philosopher(void * ph_num )
{
int num=(int)ph_num;
while(1)
{
if(num < THREADS - 1)
{
thinking(num);
//pick up right chopstick
printf("[%i] picking %i\n", num, (num + 1) % THREADS);
sem_wait(&chopstick[(num + 1) % THREADS]);
printf("[%i] picked %i\n", num, (num + 1) % THREADS);
//to make deadlocks absolutly happen, wait 1 sec then pickup left chopstick
//sleep(1);
//pick up left chopstick
printf("[%i] picking %i\n", num, num);
sem_wait(&chopstick[num]);
printf("[%i] picked %i\n", num, num);
eat(num);
//put down right chopstick
printf("[%i] put %i\n", num, (num + 1) % THREADS);
sem_post(&chopstick[(num + 1) % THREADS]);
//put down left chopstick
printf("[%i] put %i\n", num, num);
sem_post(&chopstick[num]);
}
else // last one pick left chopstick first, instead of right one to avoid cyclic wait
{
thinking(num);
//pick up left chopstick
printf("[%i] picking %i\n", num, num);
sem_wait(&chopstick[num]);
printf("[%i] picked %i\n", num, num);
//to make deadlocks absolutly happen, wait 1 sec then pickup left chopstick
//sleep(1);
//pick up right chopstick
printf("[%i] picking %i\n", num, num+1);
sem_wait(&chopstick[(num + 1) % THREADS]);
printf("[%i] picked %i\n", num, num+1);
eat(num);
//put down left chopstick
printf("[%i] put %i\n", num, num);
sem_post(&chopstick[num]);
//put down right chopstick
printf("[%i] put %i\n", num, (num + 1) % THREADS);
sem_post(&chopstick[(num + 1) % THREADS]);
}
}
pthread_exit((void *)num);
}
int main ()
{
for(int i = 0; i < THREADS; i++)
{
sem_init(&chopstick[i],0,1);
}
pthread_t threads[THREADS];
for(int i = 0; i < THREADS; i++)
pthread_create(&threads[i], NULL, philosopher, (void *)i);
for(int i = 0; i < THREADS; i++)
pthread_join(threads[i],NULL);
return 0;
}
我正在尝试为哲学家就餐问题（有五个哲学家）实施一个简单的解决方案，我的解决方案基于以下逻辑：
sem_t S[philosophers_number]
for each philosopher
{
while(TRUE)
{
if(current philosopher number != last philosopher)
{
thinking()
//i is number of current philosopher
sem_wait(take_chopstick(S[(i+1) % philosophers_number])) // right chopstick
sem_wait(take_chopstick(S[i])) // left chopstick
eat()
sem_post(put_chopstick(S[(i+1) % philosophers_number]))
sem_post(put_chopstick(S[i]))
}
else
{
thinking()
//i is number of current philosopher
sem_wait(take_chopstick(S[i])) // left chopstick
sem_wait(take_chopstick(S[(i+1) % philosophers_number])) // right chopstick
eat()
sem_post(put_chopstick(S[i]))
sem_post(put_chopstick(S[(i+1) % philosophers_number]))
}
}
每个哲学家首先思考不到三秒钟
然后如果右边的筷子可用哲学家就拿它，如果也有左边的筷子哲学家也会拿那个并开始吃不到三秒钟
然后哲学家会放下筷子让其他哲学家可以使用
为了避免循环等待，最后一个哲学家我会先拿左边的筷子再拿右边的筷子，重复同样的过程
下面是我根据这个逻辑实现的代码：
#include <stdio.h>
#include <string.h>
#include <pthread.h>
#include <semaphore.h>
#include <unistd.h>
#include <stdlib.h>
#define THREADS 5
sem_t chopstick[THREADS];
void thinking(int ph_num)
{
printf("philosopher %d is thinking\n", ph_num);
int t = rand() % 3;
sleep(t);// up to 3 secs thinking
}
void eat(int ph_num)
{
printf("philosopher %d is eating\n", ph_num);
int t = rand() % 3;
sleep(t);// up to 3 secs eating
}
void *philosopher(void * ph_num )
{
int num=(int)ph_num;
while(1)
{
if(num < THREADS - 1)
{
thinking(num);
//pick up right chopstick
sem_wait(&chopstick[(num + 1) % THREADS]);
//to make deadlocks absolutly happen, wait 1 sec then pickup left chopstick
sleep(1);
//pick up left chopstick
sem_wait(&chopstick[num]);
eat(num);
//put down right chopstick
sem_post(&chopstick[(num + 1) % THREADS]);
//put down left chopstick
sem_post(&chopstick[num]);
}
else // last one pick left chopstick first, instead of right one to avoid cyclic wait
{
thinking(num);
//pick up left chopstick
sem_wait(&chopstick[num]);
//to make deadlocks absolutly happen, wait 1 sec then pickup left chopstick
sleep(1);
//pick up right chopstick
sem_wait(&chopstick[(num + 1) % THREADS]);
eat(num);
//put down left chopstick
sem_post(&chopstick[num]);
//put down right chopstick
sem_post(&chopstick[(num + 1) % THREADS]);
}
}
pthread_exit((void *)num);
}
int main ()
{
for(int i = 0; i < THREADS; i++)
{
sem_init(&chopstick[i],0,1);
}
pthread_t threads[THREADS];
for(int i = 0; i < THREADS; i++)
pthread_create(&threads[i], NULL, philosopher, (void *)i);
for(int i = 0; i < THREADS; i++)
pthread_join(threads[i],NULL);
return 0;
}
但是在调试这段代码的过程中发生了一个问题，其中 chopstick[i] 在 sem_wait(&chopstick[num]) 之前 0 而不是阻塞当前线程，直到筷子可用 sem_wait() 继续, 所以一位哲学家开始吃饭时没有真正的筷子。
谁能帮我看看我的问题出在哪里？
你的实现是正确的，问题出在调试方法上。如果你使用 gdb，你将只在一个线程上停止，而线程的其余部分将继续执行，所以在你检查信号量和你进入下一行的时间之间，其他线程将推进执行并可以更改您检查过的值。
为了有效调试线程，您需要确保只有当前观察到的线程被调度，其余线程被阻塞。为此，您需要在线程停止后更改 scheduler-locking。您可以将其设置为 on 或 step，具体取决于您是希望线程完全停止，还是仅在单步操作期间停止（有关详细信息，请参阅 help set scheduler-locking）。
一旦线程被锁定，您可以使用info threads 来检查当时其余线程在做什么。可以用thread <<n>>换成第n个线程，用where查看线程栈。
这里是调度程序设置为 step 的示例。您可以看到只有一个线程在 next 命令上取得进展。
(gdb) b 37
Breakpoint 1 at 0x1388: file test003.c, line 37.
(gdb) r
Starting program: /home/jordan/Development/tmptest/a.out
[Thread debugging using libthread_db enabled]
Using host libthread_db library "/lib/x86_64-linux-gnu/libthread_db.so.1".
[New Thread 0x7ffff7d90700 (LWP 4002538)]
philosopher 0 is thinking
[New Thread 0x7ffff758f700 (LWP 4002539)]
philosopher 1 is thinking
[New Thread 0x7ffff6d8e700 (LWP 4002540)]
philosopher 2 is thinking
[2] picking 3
[New Thread 0x7ffff658d700 (LWP 4002541)]
[Switching to Thread 0x7ffff6d8e700 (LWP 4002540)]
Thread 4 "a.out" hit Breakpoint 1, philosopher (ph_num=0x2) at test003.c:37
37 sem_wait(&chopstick[(num + 1) % THREADS]);
(gdb) set scheduler-locking step
(gdb) info threads
Id Target Id Frame
1 Thread 0x7ffff7d91740 (LWP 4002534) "a.out" clone () at ../sysdeps/unix/sysv/linux/x86_64/clone.S:78
2 Thread 0x7ffff7d90700 (LWP 4002538) "a.out" 0x00007ffff7e743bf in __GI___clock_nanosleep (clock_id=clock_id@entry=0, flags=flags@entry=0,
req=req@entry=0x7ffff7d8fe60, rem=rem@entry=0x7ffff7d8fe60) at ../sysdeps/unix/sysv/linux/clock_nanosleep.c:78
3 Thread 0x7ffff758f700 (LWP 4002539) "a.out" 0x00007ffff7e743bf in __GI___clock_nanosleep (clock_id=clock_id@entry=0, flags=flags@entry=0,
req=req@entry=0x7ffff758ee60, rem=rem@entry=0x7ffff758ee60) at ../sysdeps/unix/sysv/linux/clock_nanosleep.c:78
* 4 Thread 0x7ffff6d8e700 (LWP 4002540) "a.out" philosopher (ph_num=0x2) at test003.c:37
5 Thread 0x7ffff658d700 (LWP 4002541) "a.out" clone () at ../sysdeps/unix/sysv/linux/x86_64/clone.S:78
(gdb) n
38 printf("[%i] picked %i\n", num, (num + 1) % THREADS);
(gdb) info threads
Id Target Id Frame
1 Thread 0x7ffff7d91740 (LWP 4002534) "a.out" clone () at ../sysdeps/unix/sysv/linux/x86_64/clone.S:78
2 Thread 0x7ffff7d90700 (LWP 4002538) "a.out" 0x00007ffff7e743bf in __GI___clock_nanosleep (clock_id=clock_id@entry=0, flags=flags@entry=0,
req=req@entry=0x7ffff7d8fe60, rem=rem@entry=0x7ffff7d8fe60) at ../sysdeps/unix/sysv/linux/clock_nanosleep.c:78
3 Thread 0x7ffff758f700 (LWP 4002539) "a.out" 0x00007ffff7e743bf in __GI___clock_nanosleep (clock_id=clock_id@entry=0, flags=flags@entry=0,
req=req@entry=0x7ffff758ee60, rem=rem@entry=0x7ffff758ee60) at ../sysdeps/unix/sysv/linux/clock_nanosleep.c:78
* 4 Thread 0x7ffff6d8e700 (LWP 4002540) "a.out" philosopher (ph_num=0x2) at test003.c:38
5 Thread 0x7ffff658d700 (LWP 4002541) "a.out" clone () at ../sysdeps/unix/sysv/linux/x86_64/clone.S:78
如你所见，执行next后，我一直在同一个线程，其他线程没有进行。
我已经修改了代码以使正在发生的事情更加可见，这是我使用的代码：
#include <stdio.h>
#include <string.h>
#include <pthread.h>
#include <semaphore.h>
#include <unistd.h>
#include <stdlib.h>
#define THREADS 5
sem_t chopstick[THREADS];
void thinking(int ph_num)
{
printf("philosopher %d is thinking\n", ph_num);
int t = rand() % 3;
sleep(t);// up to 3 secs thinking
}
void eat(int ph_num)
{
printf("philosopher %d is eating\n", ph_num);
int t = rand() % 3;
sleep(t);// up to 3 secs eating
}
void *philosopher(void * ph_num )
{
int num=(int)ph_num;
while(1)
{
if(num < THREADS - 1)
{
thinking(num);
//pick up right chopstick
printf("[%i] picking %i\n", num, (num + 1) % THREADS);
sem_wait(&chopstick[(num + 1) % THREADS]);
printf("[%i] picked %i\n", num, (num + 1) % THREADS);
//to make deadlocks absolutly happen, wait 1 sec then pickup left chopstick
//sleep(1);
//pick up left chopstick
printf("[%i] picking %i\n", num, num);
sem_wait(&chopstick[num]);
printf("[%i] picked %i\n", num, num);
eat(num);
//put down right chopstick
printf("[%i] put %i\n", num, (num + 1) % THREADS);
sem_post(&chopstick[(num + 1) % THREADS]);
//put down left chopstick
printf("[%i] put %i\n", num, num);
sem_post(&chopstick[num]);
}
else // last one pick left chopstick first, instead of right one to avoid cyclic wait
{
thinking(num);
//pick up left chopstick
printf("[%i] picking %i\n", num, num);
sem_wait(&chopstick[num]);
printf("[%i] picked %i\n", num, num);
//to make deadlocks absolutly happen, wait 1 sec then pickup left chopstick
//sleep(1);
//pick up right chopstick
printf("[%i] picking %i\n", num, num+1);
sem_wait(&chopstick[(num + 1) % THREADS]);
printf("[%i] picked %i\n", num, num+1);
eat(num);
//put down left chopstick
printf("[%i] put %i\n", num, num);
sem_post(&chopstick[num]);
//put down right chopstick
printf("[%i] put %i\n", num, (num + 1) % THREADS);
sem_post(&chopstick[(num + 1) % THREADS]);
}
}
pthread_exit((void *)num);
}
int main ()
{
for(int i = 0; i < THREADS; i++)
{
sem_init(&chopstick[i],0,1);
}
pthread_t threads[THREADS];
for(int i = 0; i < THREADS; i++)
pthread_create(&threads[i], NULL, philosopher, (void *)i);
for(int i = 0; i < THREADS; i++)
pthread_join(threads[i],NULL);
return 0;
}