我正在尝试使用pthread互斥变量和屏障来同步我的程序的输出,但它并不按我希望的方式工作。每个线程都会看到每20个值的最终值(来自for循环),但是我试图让它们都达到相同的最终值(如果使用5个线程,它们都应该看到100个为最终值,4线程,80等)通过在C中使用互斥锁和屏障进行线程同步
#include <pthread.h>
#include <stdio.h>
#include <stdlib.h>
pthread_mutex_t mutex1 = PTHREAD_MUTEX_INITIALIZER;
int SharedVariable =0;
void *SimpleThread(void *args)
{
int num,val,rc;
int which =(int)args;
rc = pthread_mutex_lock(&mutex1);
for(num=0; num<20; num++){
#ifdef PTHREAD_SYNC
if(random() > RAND_MAX/2)
usleep(10);
#endif
//pthread_mutex_lock(&mutex1);
val = SharedVariable;
printf("*** thread %d sees value %d\n", which, val);
//pthread_mutex_lock(&mutex1);
SharedVariable = val+1;
pthread_mutex_unlock(&mutex1);
}
val=SharedVariable;
printf("Thread %d sees final value %d\n", which, val);
//pthread_mutex_destroy(&mutex1);
//pthread_exit((void*) 0);
//pthread_mutex_unlock(&mutex1);
}
int main (int argc, char *argv[])
{
if(atoi(argv[1]) > 0){
int num_threads = atoi(argv[1]);
//pthread_mutex_init(&mutex1, NULL);
pthread_t threads[num_threads];
int rc;
long t;
rc = pthread_mutex_lock(&mutex1);
for(t=0; t< num_threads; t++){
printf("In main: creating thread %ld\n", t);
rc = pthread_create(&threads[t], NULL, SimpleThread, (void*)t);
if (rc){
printf("ERROR; return code from pthread_create() is %d\n", rc);
exit(-1);
}
//pthread_join(thread1);
}
rc= pthread_mutex_unlock(&mutex1);
}
else{
printf("ERROR: The parameter should be a valid positive number.");
exit(-1);
}
pthread_mutex_destroy(&mutex1);
pthread_exit(NULL);
}
任何建议或帮助,非常感谢! 感谢先进!
时,它读取的argv [1],并与开始一个数字会读取第一个数字并忽略其余数字,如“3sdfsdf”将创建3个线程。任何想法如何解决这个问题? – randomizertech
@ fgualda87:使用'strtol()'而不是'atoi()' - 它会告诉你转换后的数字后面的第一个字符,你可以用它来查看整个字符串是否有效。如果这不够清楚,请将此作为另一个问题。 – caf