拋出異常的Java 8 Lambda函數(shù)？我知道如何創(chuàng)建對(duì)具有String參數(shù)并返回int它是：Function<String, Integer>但是，如果函數(shù)拋出異常，則此操作不起作用，例如，它被定義為：Integer myMethod(String s) throws IOException我將如何定義這個(gè)引用？
3 回答
慕后森
TA貢獻(xiàn)1802條經(jīng)驗(yàn) 獲得超5個(gè)贊
如果是您的代碼，那么定義您自己的函數(shù)接口來(lái)聲明選中的異常： @FunctionalInterfacepublic interface CheckedFunction<T, R> { R apply(T t) throws IOException;}并使用它： void foo (CheckedFunction f) { ... }否則，包裝 Integer myMethod(String s)在不聲明選中異常的方法中： public Integer myWrappedMethod(String s) { try { return myMethod(s); } catch(IOException e) { throw new UncheckedIOException(e); }}然后： Function<String, Integer> f = (String t) -> myWrappedMethod(t);
或： Function<String, Integer> f = (String t) -> { try { return myMethod(t); } catch(IOException e) { throw new UncheckedIOException(e); } };
慕的地10843
TA貢獻(xiàn)1785條經(jīng)驗(yàn) 獲得超8個(gè)贊
ConsumerFunction
Consumer):
@FunctionalInterfacepublic interface ThrowingConsumer<T> extends Consumer<T> {
@Override
default void accept(final T elem) {
try {
acceptThrows(elem);
} catch (final Exception e) {
// Implement your own exception handling logic here..
// For example:
System.out.println("handling an exception...");
// Or ...
throw new RuntimeException(e);
}
}
void acceptThrows(T elem) throws Exception;}final List<String> list = Arrays.asList("A", "B", "C");forEach
final Consumer<String> consumer = aps -> {
try {
// maybe some other code here...
throw new Exception("asdas");
} catch (final Exception ex) {
System.out.println("handling an exception...");
}};list.forEach(consumer);final ThrowingConsumer<String> throwingConsumer = aps -> {
// maybe some other code here...
throw new Exception("asdas");};list.forEach(throwingConsumer);list.forEach((ThrowingConsumer<String>) aps -> {
// maybe some other code here...
throw new Exception("asda");});更新System.out...throw RuntimeException
list.forEach(Errors.rethrow().wrap(c -> somethingThatThrows(c)));
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