如何使用java8 lambda表达式抛出自定义检查的异常?

参见英文答案 > Java 8: Lambda-Streams, Filter by Method with Exception                                    13个
我有下面的代码.

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?

最佳答案
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.

转载注明原文:如何使用java8 lambda表达式抛出自定义检查的异常? - 代码日志