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我有一个程序与4个线程,最后我应该打印线程和管道(2个周期)阶段。喜欢的东西:如何让4个线程互相交互?
Thread 2: Stage 1 and 2
Thread 3: Stage 2 and 3
Thread 1: Stage 3 and 4
Thread 4: Stage 4 and 5
但我不知道如何做到这一点柜台阶段,因为我在做什么,我不能证明什么,但1级和2为每个线程,而不是1和2 ,2和3 ...
#include <stdio.h>
#include <pthread.h>
#include <time.h>
#include <stdlib.h>
#include <unistd.h>
pthread_mutex_t mutex = PTHREAD_MUTEX_INITIALIZER;
void delay (int miliseconds){
long pause;
clock_t now,then;
pause = miliseconds*(CLOCKS_PER_SEC/1000);
now = then = clock();
while((now-then) < pause)
now = clock();
}
int print(int n, int parar){
if (n == 5) {
printf("\nEstágio 4 e 5");
} else {
printf("Estágio %i e %i\n",n, n+1);
if (parar == 2) {
return (n+1, parar+0);
}
return print(n+1, parar+1);
}
}
void thread_cont(void *arg){
int *pvalor;
pvalor=arg;
pthread_mutex_lock(&mutex);
printf ("\n---Thread %i--- \n", *pvalor);
int a = print(1, 1);
delay(2000);
pthread_mutex_unlock(&mutex);
/*
if (*pvalor == 1){
printf ("Estágio 1 e 2");
}
if (*pvalor == 2){
printf ("Estágio 2 e 3");
}
if (*pvalor == 3){
printf ("Estágio 3 e 4");
}
if (*pvalor == 4){
printf ("Estágio 4 e 5");
}
*/
}
int main() {
pthread_t id1;
int offset1 = 1;
pthread_create(&id1, NULL, thread_cont, &offset1);
pthread_t id2;
int offset2 = 2;
pthread_create(&id2, NULL, thread_cont, &offset2);
pthread_t id3;
int offset3 = 3;
pthread_create(&id3, NULL, thread_cont, &offset3);
pthread_t id4;
int offset4 = 4;
pthread_create(&id4, NULL, thread_cont, &offset4);
pthread_join(id1, NULL);
pthread_join(id2, NULL);
pthread_join(id3, NULL);
pthread_join(id4, NULL);
printf("\n");
return 0;
}
因为每个线程调用'打印(1,1)',它是很难理解为什么你期望不同的行为。也许你应该使用'print(* pvalor,* pvalor + 1)'?我不清楚'print()'函数真的应该做什么,以及它为什么递归地调用它自己。另外,'return(n + 1,parar + 0);'和'return parar没有什么不同;'你也打算在那里调用'print()'吗?为什么'+ 0'? –