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c - pthread_mutex_timedlock() 过早退出而不等待超时

转载 作者:IT王子 更新时间:2023-10-29 01:11:06 27 4
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我想保护一个函数免受多线程访问。为此,我使用了 pthread_mutex_t 互斥体。我尝试在函数的开头锁定它,然后执行该函数,然后再次释放它。如果互斥锁正在使用中,它应该最多等待 60 秒才能可用。如果之后它仍然不可用,则该函数应该失败。

我遇到的问题是 pthread_mutex_timedlock 似乎完全忽略了我给它的超时值。虽然我指定了 60 秒的超时,但如果锁定,该函数会立即返回并返回错误代码 ETIMEDOUT -- 而无需实际等待。

这是一个重现问题的最小示例。在这种情况下,使用递归或非递归互斥并不重要,因为我不会尝试从同一线程多次锁定它们。

#include <stdlib.h>
#include <stdio.h>
#include <stdint.h>
#include <stdbool.h>
#include <string.h>
#include <stddef.h>
#include <unistd.h>
#include <time.h>
#include <errno.h>
#include <pthread.h>

pthread_mutex_t lock; /* exclusive lock */

//do some work to keep the processor busy..
int wut() {
int x = 0;
for(int i=0; i < 1024*1024*1024; i++)
x += 1;
return x;
}

void InitMutex(){
/*pthread_mutexattr_t Attr;
pthread_mutexattr_init(&Attr);
pthread_mutexattr_settype(&Attr, PTHREAD_MUTEX_RECURSIVE);
pthread_mutex_init(&lock, &Attr);*/
pthread_mutex_init(&lock, NULL);
}

//lock mutex, wait at maximum 60 seconds, return sucesss
int LockMutex() {
struct timespec timeoutTime;
timeoutTime.tv_nsec = 0;
timeoutTime.tv_sec = 60;
printf("Nanoseconds: %lu, seconds %lu\n", timeoutTime.tv_nsec, timeoutTime.tv_sec);
int retVal = pthread_mutex_timedlock(&lock, &timeoutTime);
printf("pthread_mutex_timedlock(): %d\n", retVal);
if(retVal != 0) {
const char* errVal = NULL;
switch(retVal) {
case EINVAL: errVal = "EINVAL"; break;
case EAGAIN: errVal = "EAGAIN"; break;
case ETIMEDOUT: errVal = "ETIMEDOUT"; break;
case EDEADLK: errVal = "EDEADLK"; break;
default: errVal = "unknown.."; break;
}
printf("Error taking lock in thread %lu: %s (%s)\n", pthread_self(), errVal , strerror(retVal));
}
return retVal == 0; //indicate success/failure
}

void UnlockMutex() {
pthread_mutex_unlock(&lock);
}

void TestLockNative() {
uint64_t thread_id = pthread_self();
printf("Trying to take lock in thread %lu.\n", thread_id);
int ret = LockMutex();
printf("Got lock in thread %lu. sucess=%d\n", thread_id, ret);
wut();
printf("Giving up lock now from thread %lu.\n", thread_id);
UnlockMutex();

}

void* test_thread(void* arg) {
//TestLock();
TestLockNative();
return NULL;
}

int main() {
InitMutex();
//create two threads which will try to access the protected function at once
pthread_t t1, t2;
pthread_create(&t1, NULL, &test_thread, NULL);
pthread_create(&t2, NULL, &test_thread, NULL);

//wait for threads to end
pthread_join(t1, NULL);
pthread_join(t2, NULL);

return 0;
}

程序的输出例如:

Trying to take lock in thread 139845914396416.
Nanoseconds: 0, seconds 6000
pthread_mutex_timedlock(): 0
Got lock in thread 139845914396416. sucess=1
Trying to take lock in thread 139845906003712.
Nanoseconds: 0, seconds 6000
pthread_mutex_timedlock(): 110
Error taking lock in thread 139845906003712: ETIMEDOUT (Connection timed out) [<-- this occurs immediately, not after 60 seconds]
Got lock in thread 139845906003712. sucess=0
Giving up lock now from thread 139845906003712.

使用 gcc -o test test.c -lpthread 编译应该可以。

那么,有谁知道这里发生了什么以及为什么 pthread_mutex_timedlock() 忽略了我的超时值?它不表现 the way it is documented完全没有。

我使用的是 Ubuntu 16.04.2 LTS 系统,使用 gcc 编译。

最佳答案

pthread_mutex_timedlock 的手册页说:

The timeout shall expire when the absolute time specified by abstime passes, as measured by the clock on which timeouts are based

因此,使用实时来指定您的超时值:

int LockMutex() {
struct timespec timeoutTime;
clock_gettime(CLOCK_REALTIME, &timeoutTime);
timeoutTime.tv_sec += 60;

int retVal = pthread_mutex_timedlock(&lock, &timeoutTime);
....

关于c - pthread_mutex_timedlock() 过早退出而不等待超时,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46365448/

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