为什么在再次调用 Thread.start 时会发生 IllegalThreadStateException

multithreading java

2022-09-02 10:24:24

public class SieveGenerator{

static int N = 50;
public static void main(String args[]){

    int cores = Runtime.getRuntime().availableProcessors();

    int f[] = new int[N];

    //fill array with 0,1,2...f.length
    for(int j=0;j<f.length;j++){
        f[j]=j;
    }

    f[0]=0;f[1]=0;//eliminate these cases

    int p=2;

    removeNonPrime []t = new removeNonPrime[cores];

    for(int i = 0; i < cores; i++){
        t[i] = new removeNonPrime(f,p);
    }

    while(p <= (int)(Math.sqrt(N))){
        t[p%cores].start();//problem here because you cannot start a thread which has already started(IllegalThreadStateException)
        try{
            t[p%cores].join();
        }catch(Exception e){}
        //get the next prime
        p++;
        while(p<=(int)(Math.sqrt(N))&&f[p]==0)p++;
    }


    //count primes
    int total = 0;
    System.out.println();

    for(int j=0; j<f.length;j++){
        if(f[j]!=0){
            total++;
        }
    }
    System.out.printf("Number of primes up to %d = %d",f.length,total);
}
}


class removeNonPrime extends Thread{
int k;
int arr[];

public removeNonPrime(int arr[], int k){
    this.arr = arr;
    this.k = k;
}

public void run(){
    int j = k*k;
    while(j<arr.length){
        if(arr[j]%k == 0)arr[j]=0;
        j=j+arr[k];

    }
}
}

嗨，当我运行我的代码时，我得到了一个，我认为这是因为我正试图启动一个已经启动的线程。那么，我该如何每次都杀死或停止线程，以解决此问题呢？IllegalThreadStateException

答案 1

我怎么能每次都杀死或停止线程，以解决这个问题？

答案是，你不能。一旦启动，可能无法重新启动。这在 javadoc 中清楚地记录在中。相反，你真正想做的是每次你在循环中出现时的实例。ThreadThreadnewRemoveNonPrime

您的代码中还有其他一些问题。首先，在再次使用它之前，您需要递增：p

for(int i = 0; i < cores; i++){
    t[i] = new removeNonPrime(f,p); //<--- BUG, always using p=2 means only multiples of 2 are cleared
}

其次，您可能是多线程的，但您不是并发的。您拥有的代码基本上一次只允许一个线程运行：

while(p <= (int)(Math.sqrt(N))){
    t[p%cores].start();//
    try{
        t[p%cores].join(); //<--- BUG, only the thread which was just started can be running now
    }catch(Exception e){}
    //get the next prime
    p++;
    while(p<=(int)(Math.sqrt(N))&&f[p]==0)p++;
}

只是我的0.02美元，但你试图做的事情可能有效，但是选择下一个最小素数的逻辑并不总是选择一个素数，例如，如果其他线程之一尚未处理数组的该部分。

以下是使用执行器服务的方法，您必须填写一些空白（...）：

/* A queue to trick the executor into blocking until a Thread is available when offer is called */
public class SpecialSyncQueue<E> extends SynchronousQueue<E> {
    @Override
    public boolean offer(E e) {
        try {
            put(e);
            return true;
        } catch (InterruptedException ex) {
            Thread.currentThread().interrupt();
            return false;
        }
    }
}

ExecutorService executor = new ThreadPoolExecutor(cores, cores, new SpecialSyncQueue(), ...);
void pruneNonPrimes() {
    //...
    while(p <= (int)(Math.sqrt(N))) {
        executor.execute(new RemoveNonPrime(f, p));
        //get the next prime
        p++;
        while(p<=(int)(Math.sqrt(N))&&f[p]==0)p++;
    }


    //count primes
    int total = 0;
    System.out.println();

    for(int j=0; j<f.length;j++){
        if(f[j]!=0){
            total++;
        }
    }
    System.out.printf("Number of primes up to %d = %d",f.length,total);
}



class RemoveNonPrime extends Runnable {
    int k;
    int arr[];

    public RemoveNonPrime(int arr[], int k){
        this.arr = arr;
        this.k = k;
    }

    public void run(){
        int j = k*k;
        while(j<arr.length){
            if(arr[j]%k == 0)arr[j]=0;
            j+=k;
        }
    }
}

答案 2

您可以改为实现Runnable并使用new Thread（$Runnable这里）.start（）或使用ExecutorService来重用线程。