0

我的代码看起来像这样

public class ListAnalyzer {
public static List<String> base;

public static void addOfflineToBase(String offline) {
    base.add(offline);
}

public static List<String> prepareArrayList() throws IOException {
    base = Files.readAllLines(Paths.get("/Users/noname/Desktop/test.txt").toAbsolutePath().normalize());
    List<String> part = base.subList(0, 200);
    base.removeAll(part);
    saveBase();
    return part;
}

当我尝试运行它时 - 我看到：

Unable to evaluate the expression Method threw 'java.util.ConcurrentModificationException' exception.

当我尝试使用 base.removeAll(part) 时，我在这里看不到冲突。有人能帮助我吗？顺便说一句：saveBase() 只是将数据写入文本文件 UPD：Stacktrace

21:59:26: Executing task 'App.main() --scan'...

> Task :compileJava UP-TO-DATE
> Task :processResources NO-SOURCE
> Task :classes UP-TO-DATE

> Task :App.main() FAILED
2 actionable tasks: 1 executed, 1 up-to-date

Publishing a build scan to scans.gradle.com requires accepting the Gradle Terms of Service defined at https://gradle.com/terms-of-service. Do you accept these terms? [yes, no] Exception in thread "main" java.util.ConcurrentModificationException
 at java.base/java.util.ArrayList$SubList.checkForComodification(ArrayList.java:1445)
 at java.base/java.util.ArrayList$SubList.size(ArrayList.java:1185)
 at java.base/java.util.AbstractCollection.isEmpty(AbstractCollection.java:87)
 at com.lukaspradel.steamapi.webapi.request.GetPlayerSummariesRequest$GetPlayerSummariesRequestBuilder.<init>(GetPlayerSummariesRequest.java:41)
 at hexlet.code.App.getOnlineIds(App.java:39)
 at hexlet.code.App.start(App.java:21)
 at hexlet.code.App.main(App.java:16)

FAILURE: Build failed with an exception.

* What went wrong:
Execution failed for task ':App.main()'.
> Process 'command '/Library/Java/JavaVirtualMachines/zulu-11.jdk/Contents/Home/bin/java'' finished with non-zero exit value 1

* Try:
Run with --stacktrace option to get the stack trace. Run with --info or --debug option to get more log output. Run with --scan to get full insights.

* Get more help at https://help.gradle.org

BUILD FAILED in 337ms

```lang-none
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2 回答 2

0

我需要了解如何安全地从base.

有两种方法：

  1. 添加一些应用程序级别的锁定，这样您就不会尝试从列表中删除字符串，而其他东西正在迭代列表。

  2. 使用并发集合类型而不是List. （不清楚哪种集合类型最合适。没有合适的并发列表类型；即支持高效更新的类型。）并发集合类型在java.util.concurrent 包中定义。

于 2021-08-15T15:42:51.093 回答
0

我想问题来自以下几行：

List<String> part = base.subList(0, 200);
base.removeAll(part);

列表part实际上是由列表支持的base，因此更改base会反映在part. 这可能会导致这样的异常。

要确认这一点，请尝试：

List<String> part = new ArrayList<>(base.subList(0, 200));
base.removeAll(part);

另一种可能性，它做一些不同的事情，但与你的相似：

 base.subList(0, 200).clear();

这将删除 0 到 200 范围内的所有元素，前提是所有元素都是唯一的。

于 2021-08-20T10:18:41.903 回答