我正在阅读一些文档并尝试一些发出 futex
的代码示例。 Linux 中的系统调用。我读到如果 thread_a
获得了互斥锁使用 FUTEX_LOCK_PI
,并说如果另一个线程 thread_b
试图获得相同的,后者( thread_b
)在内核中被阻止。
“在内核中阻塞”究竟是什么意思?是不是执行了thread_b
在用户空间不恢复?在以下示例中,第二个线程似乎也在系统调用之后发出 printf。这不是表明它没有在内核中被阻塞吗?我错过了什么?
/*
* This example demonstrates that when futex_lock_pi is called twice, the
* second call is blocked inside the kernel.
*/
#include <stdio.h>
#include <unistd.h>
#include <pthread.h>
int mutex = 0;
#define __NR_futex 240
#define FUTEX_WAIT 0
#define FUTEX_WAKE 1
#define FUTEX_FD 2
#define FUTEX_REQUEUE 3
#define FUTEX_CMP_REQUEUE 4
#define FUTEX_WAKE_OP 5
#define FUTEX_LOCK_PI 6
#define FUTEX_UNLOCK_PI 7
#define FUTEX_TRYLOCK_PI 8
#define FUTEX_WAIT_REQUEUE_PI 11
#define FUTEX_CMP_REQUEUE_PI 12
void *thread(void *arg) {
int ret = 0;
pid_t tid = gettid();
printf("Entering thread[%d]\n", tid);
ret = syscall(__NR_futex, &mutex, FUTEX_LOCK_PI, 1, NULL, NULL, 0);
printf("Thread %d returned from kernel\n", tid);
printf("Value of mutex=%d\n", mutex);
printf("Return value from syscall=%d\n", ret);
}
int main(int argc, const char *argv[]) {
pthread_t t1, t2;
if (pthread_create(&t1, NULL, thread, NULL)) {
printf("Could not create thread 1\n");
return -1;
}
if (pthread_create(&t2, NULL, thread, NULL)) {
printf("Could not create thread 2\n");
return -1;
}
// Loop infinitely
while ( 1 ) { }
return 0;
}
输出可以在下面找到:-
Entering thread[952]
Thread 952 returned from kernel
Entering thread[951]
Value of mutex=-2147482696
Return value from syscall=0
Thread 951 returned from kernel
Value of mutex=-1073740873
Return value from syscall=0
最佳答案
在内核中阻塞意味着线程处于 sleep 状态,直到事件将其唤醒,在您的情况下,事件是可用的互斥锁。
我改变了一点你的代码来说明:
#include <stdio.h>
#include <unistd.h>
#include <pthread.h>
#include <sys/syscall.h>
static pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
void *thread(void *unused) {
printf("Entering thread %d\n", syscall(SYS_gettid));
pthread_mutex_lock(&mutex);
sleep(2);
pthread_mutex_unlock(&mutex);
return NULL;
}
int main(int argc, const char *argv[]) {
pthread_t t1, t2;
printf("using mutex @%p\n", &mutex);
if (pthread_create(&t1, NULL, thread, &t1)) {
printf("Could not create thread 1\n");
return -1;
}
if (pthread_create(&t2, NULL, thread, &t2)) {
printf("Could not create thread 2\n");
return -1;
}
pthread_join(t1, NULL);
pthread_join(t2, NULL);
return 0;
}
代码基本相同,但我明确使用了 pthread 库的互斥锁。
如果我用 strace 运行这段代码,我会得到:
using mutex @0x6010a0
Process 19688 attached
Entering thread 19688
Process 19689 attached
Entering thread 19689
[pid 19689] futex(0x6010a0, FUTEX_WAIT_PRIVATE, 2, NULL <unfinished ...>
[pid 19688] futex(0x6010a0, FUTEX_WAKE_PRIVATE, 1) = 1
[pid 19689] <... futex resumed> ) = 0
[pid 19688] +++ exited with 0 +++
[pid 19689] futex(0x6010a0, FUTEX_WAKE_PRIVATE, 1) = 0
[pid 19689] +++ exited with 0 +++
+++ exited with 0 +++
您可能会注意到,我们没有看到第一个线程使用互斥锁,但是我们可以看到下一个线程在等待 futex (FUTEX_WAIT_PRIVATE)。这是因为在获取互斥锁时不会调用 futex。
但是你可以看到第一个线程(这里的 id 为 19688)最终调用了 futex(FUTEX_WAKE_PRIVATE),它告诉内核在 utex 空闲的时候唤醒另一个线程。
你可能已经注意到第一个调用
[pid 19689] futex(0x6010a0, FUTEX_WAIT_PRIVATE, 2, NULL <unfinished ...>
未完成,这意味着进程被挂起,等待内核完成工作并交还手。然后
[pid 19689] <... futex resumed> ) = 0
调用最终完成的草图(显然是因为互斥锁被释放了)
关于c - 内核中的 Futex 和阻塞,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33208589/