当捕获实际上没有捕获任何东西

Question

由于最近存储在数据库中的数据不好,导致程序崩溃.这让我感到困惑,因为我觉得我有一个阻止这一点.

以下代码的目的是比较员工徽章编号并对其进行排序.如果出现错误,请返回-1并且士兵开启 - 请勿停止,因为数千个徽章编号中的一个错误:

public int compare(Employee t, Employee t1) {
    Integer returnValue = -1;
    try {
        Integer tb = Integer.parseInt(t.getBadgeNumber());
        Integer t1b = Integer.parseInt(t1.getBadgeNumber());
        returnValue = tb.compareTo(t1b);
    } catch (Exception e) {
        returnValue = -1;//useless statement, I know.
    }
    return returnValue;
}

当坏徽章编号被击中时(在这种情况下为t),我得到了一个"java.lang.IllegalArgumentException:比较方法违反了它的一般合同!" 错误而不是在catch中返回-1.

关于这里的捕获,我不明白什么？

完整的堆栈跟踪:

16-May-2018 14:28:53.496 SEVERE [http-nio-8084-exec-601] org.apache.catalina.core.StandardWrapperValve.invoke Servlet.service() for servlet [RequestServlet] in context with path [/AppearanceRequest] threw exception
 java.lang.IllegalArgumentException: Comparison method violates its general contract!
at java.util.TimSort.mergeHi(TimSort.java:868)
at java.util.TimSort.mergeAt(TimSort.java:485)
at java.util.TimSort.mergeForceCollapse(TimSort.java:426)
at java.util.TimSort.sort(TimSort.java:223)
at java.util.TimSort.sort(TimSort.java:173)
at java.util.Arrays.sort(Arrays.java:659)
at java.util.Collections.sort(Collections.java:217)
at org.bcso.com.appearancerequest.html.NotifierHTML.getHTML(NotifierHTML.java:363)
at org.bcso.com.appearancerequest.AppearanceRequestServlet.processRequest(AppearanceRequestServlet.java:96)
at org.bcso.com.appearancerequest.AppearanceRequestServlet.doGet(AppearanceRequestServlet.java:565)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:618)
at javax.servlet.http.HttpServlet.service(HttpServlet.java:725)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:301)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:52)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:239)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.netbeans.modules.web.monitor.server.MonitorFilter.doFilter(MonitorFilter.java:393)
at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:239)
at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:206)
at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:219)
at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:106)
at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:503)
at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:136)
at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:74)
at org.apache.catalina.valves.AbstractAccessLogValve.invoke(AbstractAccessLogValve.java:610)
at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:88)
at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:516)
at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1015)
at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:652)
at org.apache.coyote.http11.Http11NioProtocol$Http11ConnectionHandler.process(Http11NioProtocol.java:222)
at org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.doRun(NioEndpoint.java:1575)
at org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.run(NioEndpoint.java:1533)
at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1145)
at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:615)
at java.lang.Thread.run(Thread.java:745)

调用代码:

    List<Employee> employeeList = DatabaseUtil.getEmployees();
    Collections.sort(employeeList, new BadgeComparator());

Answer 1

异常(无论是什么)被抓住了catch (Exception e).您没有记录此异常,因此您不知道它是什么.你应该以某种方式记录它,以便你知道到底发生了什么.

返回时会出现问题-1.这允许Java的当前排序算法有时会捕获的顺序不一致的可能性.总之,返回-1上的错误意味着你在断言都a < b和b < a真实,因为异常会在两种情况下被抓住.这在逻辑上是不正确的.排序算法检测到这个并抛出IllegalArgumentException.请注意,该compare方法不在堆栈跟踪中; 这是对的呼唤Collections.sort.

除了记录异常之外,在进入程序中的比较步骤之前处理它.如果必须将字符串解析为整数,请在创建Employee对象时执行此操作,以便在进入程序中的排序步骤之前进行验证.A Comparator不应该验证数据; 它应该只比较数据.

Answer 2

说明

java.lang.IllegalArgumentException:比较方法违反了它的一般合同!

你的内心不会抛出异常try.这就是它没有被抓住的原因.例外情况来自NotifierHTML.java:363您调用Collection#sort哪个TimSort类使用的代码.然后TimSort.java:868由该TimSort#mergeHi方法抛出异常.

它告诉您,您的Comparator#compare方法实现是错误的.如其文档中所述,它违反了合同:

比较它的两个参数的顺序.返回负整数,零或正整数,因为第一个参数小于,等于或大于第二个参数.

实施者必须确保 sgn(x.compareTo(y)) == -sgn(y.compareTo(x))所有x和y.(这意味着x.compareTo(y)必须抛出异常iff y.compareTo(x)抛出异常.)

实现者还必须确保关系是可传递的:(x.compareTo(y) > 0 && y.compareTo(z) > 0)暗示x.compareTo(z) > 0.

最后,实现者必须确保这x.compareTo(y) == 0意味着sgn(x.compareTo(z)) == sgn(y.compareTo(z))对所有人而言z.

您的实施违反了其中一项要求,并且检测到了该方法.

问题的根源

问题是-1如果发生错误则返回.假设你有两个值first和second.并且至少其中一个会引发异常.

所以,如果你想比较first有second,你会得到-1:

compare(first, second) -> -1

这意味着first是小比second.但是,如果你将它与你得到的另一种方式进行比较-1:

compare(second, first) -> -1

因为两种变体都会引发异常,这会导致您的异常return -1;.但这意味着你的compare方法说:

first < second
second < first

两者同时存在,这在逻辑上是不正确的,违反了合同.

解

您需要正确定义订购中不可分类内容的位置.例如,让我们定义它总是小于任何数字.所以我们想要

text < number

如果两者都是不可解决的,我们该怎么办？我们可以说它们是平等的,我们可以比较它们的词典.让我们保持简单,并说任何两个文本被认为是平等的:

text = text

我们通过检查哪些参数是不可解析的然后返回正确的值来实现它:

@Override
public int compare(Employee first, Employee second) {
    Integer firstValue;
    Integer secondValue;
    try {
        firstValue = Integer.parseInt(first.getBadgeNumber());
    } catch (NumberFormatException e) {
        // Could not parse, set null as indicator
        firstValue = null;
    }
    try {
        secondValue = Integer.parseInt(second.getBadgeNumber());
    } catch (NumberFormatException e) {
        // Could not parse, set null as indicator
        secondValue = null;
    }

    if (firstValue == null && secondValue != null) {
        // text < number
        return -1;
    }
    if (firstValue != null && secondValue == null) {
        // number > text
        return 1;
    }
    if (firstValue == null && secondValue == null) {
        // text = text
        return 0;
    }

    // Both are numbers
    return Integer.compare(firstValue, secondValue);
}

正如评论中暗示的那样,您可以Comparator通过以下语句替换整个自定义类,该语句生成相同的Comparator:

Comparator<Employee> comp = Comparator.nullsLast(
    Comparator.comparing(e -> tryParseInteger(e.getBadgeNumber())));

与这样的tryParseInteger方法一起:

public static Integer tryParseInteger(String text) {
    try {
        return Integer.parseInt(text);
    } catch (NumberFormatException e) {
        return null;
    }
}