如何使用java8 lambda表达式抛出自定义检查的异常?
发布时间:2020-05-24 12:03:40 所属栏目:Java 来源:互联网
导读:参见英文答案 Java 8: Lambda-Streams, Filter by Method with Exception13个 我有下面的代码. private static void readStreamWithjava8() { StreamString lines = null; try
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参见英文答案 >
Java 8: Lambda-Streams,Filter by Method with Exception13个
private static void readStreamWithjava8() {
Stream<String> lines = null;
try {
lines = Files.lines(Paths.get("b.txt"),StandardCharsets.UTF_8);
lines.forEachOrdered(line -> process(line));
} catch (IOException e) {
e.printStackTrace();
} finally {
if (lines != null) {
lines.close();
}
}
}
private static void process(String line) throws MyException {
// Some process here throws the MyException
}
这里我的进程(String line)方法抛出已检查的异常,我从lambda中调用该方法.此时需要从readStreamWithjava8()方法抛出MyException而不抛出RuntimeException. 我怎么能用java8做到这一点? 解决方法简短的回答是,你做不到.这是因为forEachOrdered接受了Consumer,并且未声明Consumer.accept会抛出任何异常.解决方法是做类似的事情 List<MyException> caughtExceptions = new ArrayList<>();
lines.forEachOrdered(line -> {
try {
process(line);
} catch (MyException e) {
caughtExceptions.add(e);
}
});
if (caughtExceptions.size() > 0) {
throw caughtExceptions.get(0);
}
但是,在这些情况下,我通常会在process方法中处理异常,或者使用for-loops以旧式方式处理异常. (编辑:安卓应用网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |
