遞回方法回傳后執行代碼

我正在嘗試使用遞回方法實作重試機制。盡管該方法已經回傳了串列，但連續的代碼行也會被執行。

我確信我錯過了一些非常明顯的東西，訓練有素的眼睛可以在一秒鐘內發現。我怎樣才能重做遞回，所以我擺脫了最后一條return null陳述句。它根本不需要。

這是整個代碼：

import java.io.IOException;
import java.util.ArrayList;
import java.util.List;
import java.util.concurrent.*;

public class Retryer {

    private static final int RETRY_COUNT = 3;
    private static final long RETRY_DELAY = 2; // in seconds

    private static int tryCounter = 0;

    public Retryer() {}

    public void execute() {
        List<String> downloadedFiles;
        try {
            Callable<List<String>> c = new AsyncFileDownloader();
            downloadedFiles = c.call();
        } catch (Exception e) {
            System.out.println("Ops! Starting retry");
            ScheduledExecutorService scheduledExecutorService = Executors.newSingleThreadScheduledExecutor();
            downloadedFiles = retry(scheduledExecutorService, 0);
            System.out.println("Retry ended");
            if (downloadedFiles == null || downloadedFiles.isEmpty()) {
                System.out.println("Sorry. Will exit now");
                System.exit(0);
            }
        }

        System.out.println("Finished successfully with "   downloadedFiles.size()   " in list.");
    }

    private List<String> retry (ScheduledExecutorService ses, int retryCounter) {
        if (retryCounter == RETRY_COUNT) {
            System.out.println("Retry limit reached. Exiting");
            ses.shutdown();
            return null;
        }
        try {
            System.out.println("Retry attempt # "   (retryCounter 1)   " will start in "   RETRY_DELAY   " seconds ...");
            ScheduledFuture sf = ses.schedule(new AsyncFileDownloader(), RETRY_DELAY, TimeUnit.SECONDS);
            List<String> downloadedFiles = (List<String>) sf.get();
            ses.shutdown();
            System.out.println("In retry. success "   downloadedFiles );
            return downloadedFiles;
        } catch (Exception e) {
            System.out.println("--recursive calling");
            retry(ses,   retryCounter);
        }
        System.out.println("--returning null!!! This shouldn't happen");
        return null;
    }

    class AsyncFileDownloader implements Callable<List<String>> {
        @Override
        public List<String> call() throws Exception {
            synchronized (this) {
                tryCounter  ;
                System.out.println("--in pull");
                if (tryCounter < 3) throw new IOException("Some bad thing happened");
                List<String> retList = new ArrayList<>();
                retList.add("file1");
                retList.add("file1");
                return retList;
            }
        }
    }
}

這里的輸出：

--in pull
Ops! Starting retry
Retry attempt # 1 will start in 2 seconds ...
--in pull
--recursive calling
Retry attempt # 2 will start in 2 seconds ...
--in pull
In retry. success [file1, file1]
--returning null!!! This shouldn't happen
Retry ended
Sorry. Will exit now

uj5u.com熱心網友回復：

如果第一次執行進入 catch 塊，則遞回呼叫重試。一旦第二次呼叫回傳，您在第一次呼叫中的 catch 塊之后繼續執行代碼，導致您點擊列印“回傳 null”。

您真正想要的可能是回傳遞回呼叫的值。請記住，它return向呼叫另一個函式的函式回傳一個值，不一定是呼叫鏈中的第一個函式。

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