c - 如何在 Linux C 上同步两个线程?
问题描述
所以我想通过这段代码同步两个线程。顺便说一下,让我解释一下,我希望 thread1 从传感器读取一个值并将其写入一个名为“integers.dat”的文件。因此,thread2 的任务是将之前写在“integers.dat”上的内容发送到 gnuplot(如下面的代码所示)。所以我希望线程执行顺序如下:thread1(写),thread2(发送),thread1(写),thread2(发送),依此类推。我试图用互斥锁来做到这一点,但没有奏效。两个线程的执行总是随机的。
代码:`
#include <stdio.h>
#include <pigpio.h>
#include<sys/types.h>
#include<signal.h>
#include<unistd.h>
#include<pthread.h>
#include<semaphore.h>
#define TRIGGER 5
#define ECHO 6
void *thread1_process (void *arg);
void *thread2_process( void *arg);
double start, stop, measure;
int i, val, it=0;
FILE *fptr;
FILE *gnu ;
static pthread_mutex_t my_mutex11;
int main(int argc, char *argv[]) {
pthread_t th1, th2;
void *ret;
fptr=fopen("integers.dat", "w");
gnu = popen("gnuplot -persistent","w");
pthread_mutex_init (&my_mutex11, NULL);
pthread_create(&th1, NULL, thread1_process, NULL);
pthread_create (&th2, NULL, thread2_process, NULL);
(void)pthread_join (th1, &ret);
(void)pthread_join (th2, &ret);
}
void *thread1_process (void *arg)
{
for(int i=0; i<20; i++)
{
pthread_mutex_lock (&my_mutex11);
printf("thread1 %d \n", i);
gpioInitialise();
gpioSetMode(TRIGGER , PI_OUTPUT); // trigger
gpioSetMode(ECHO , PI_INPUT);
gpioWrite(TRIGGER, 0);
gpioSleep(PI_TIME_RELATIVE, 0, 1);
gpioWrite(TRIGGER, 1);
gpioSleep(PI_TIME_RELATIVE, 0, 10); // sleep for 0.00001 seconds
gpioWrite(TRIGGER, 0);
while (gpioRead(ECHO) == 0)
start = time_time();
while (gpioRead(ECHO) == 1)
stop = time_time();
stop=time_time();
measure = (stop-start) *17100.50;
it++;
val=measure;
fprintf(fptr, "%d %d\n", it, val);
gpioTerminate();
pthread_mutex_unlock (&my_mutex11);
}
pthread_exit(0);
}
void *thread2_process( void *arg)
{
for(int j=0; j<20; j++)
{
pthread_mutex_lock (&my_mutex11);
printf("thread2 %d \n", j);
fprintf(gnu, "%s \n","plot 'integers.dat' with linespoints lw 3");
pthread_mutex_unlock (&my_mutex11);
}
pthread_exit (0);
}`
结果
[thread1 0
thread2 0
thread2 1
thread2 2
thread2 3
thread2 4
thread2 5
thread2 6
thread2 7
thread2 8
thread2 9
thread2 10
thread2 11
thread2 12
thread2 13
thread2 14
thread2 15
thread2 16
thread2 17
thread2 18
thread2 19
thread1 1
thread1 2
Warning: empty y range [108:108], adjusting to [106.92:109.08]
libEGL warning: DRI2: failed to authenticate
thread1 3
thread1 4
thread1 5
thread1 6
thread1 7
thread1 8
thread1 9
thread1 10
thread1 11
thread1 12
thread1 13
thread1 14
thread1 15
thread1 16
thread1 17
thread1 18
thread1 19]
解决方案
除了明显的“如果一个线程总是在等待,为什么要使用两个线程?” 您需要做的是在互斥锁中使用某种“状态”变量:
// start here
#define STATE_INITIAL 0
// go here when step 1 finishes
#define STATE_STEP1 1
// go here when step 2 finishes
#define STATE_STEP2 2
pthread_mutex_t my_mutex;
int cur_state;
void wait_my_turn(int desired_state) {
pthread_mutex_lock(&my_mutex);
if (cur_state == desired_state) return;
pthread_mutex_unload(&my_mutex);
}
void finish_turn() {
++cur_state;
if (cur_state == 3) cur_state = 1;
pthread_mutex_unlock(&my_mutex);
}
// in main, initialize mutex, lock it, and set cur_state to STATE_INITIAL
// until you are ready for threads to start. Then you need to set it
// to STATE_STEP1 to allow that thread to begin.
// in your threads, begin with wait_my_turn(STATE_STEPn)
// and call finish_turn() when done
...不是一个很好的例子,但你应该了解它的要点。
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