我正在尝试为哲学家进餐问题（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的时间不到三秒钟
然后如果右边的筷子可用，哲学家会拿它，如果还有左边的筷子可用，哲学家也会拿它，开始吃不到三秒钟
然后哲学家会放下筷子，让其他哲学家使用
为了避免回圈等待，对于最后一位哲学家，我将先拿起左筷子，然后再拿起右筷子，然后继续相同的程序
这是我基于此逻辑实作的代码：
#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;
}
But during debugging this code a problem happened, where chopstick[i] was 0 before sem_wait(&chopstick[num]) instead of blocking current thread, until a chopstick is available sem_wait() carried on, so a philosopher started eating without an actual chopstick.
Can anyone help me figure out where is my problem?
uj5u.com热心网友回复：
您的实作是正确的，您的问题出在除错方法上。如果您使用gdb，您将仅在一个执行绪上停止，而执行绪的其余部分将继续执行，因此在您检查信号量的时间和您进入下一行的时间之间，其他执行绪将继续执行并可以更改您检查过的值。
为了有效地除错执行绪，您需要确保只有当前观察到的执行绪被调度而其余执行绪被阻塞。为此，您需要scheduler-locking在停止执行绪后更改。您可以将其设定为on或step，具体取决于您是希望执行绪完全停止，还是仅在单步操作期间停止（help set scheduler-locking有关更多详细信息，请参阅 参考资料）。
一旦执行绪被锁定，您就可以使用它info threads来检查其余执行绪当时在做什么。您可以使用thread <<n>>更改为第 n 个执行绪并用于where检查执行绪堆栈。
这是调度程序设定为 的示例step。您可以看到只有一个执行绪执行了该next命令。
(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
如您所见，执行 next 后，我??仍停留在同一个执行绪上，而其他执行绪没有进展。
我已经修改了代码以使正在发生的事情更加明显，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;
}
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