我知道如何创建一个有String参数并返回int的方法的引用,它是:

Function<String, Integer>

然而,如果函数抛出异常,比如它被定义为:

Integer myMethod(String s) throws IOException

我该如何定义这个引用呢?


当前回答

使用Function包装器的另一种解决方案是返回结果包装器的实例,如果一切正常,则返回成功,或者返回失败。

一些代码来澄清事情:

public interface ThrowableFunction<A, B> {
    B apply(A a) throws Exception;
}

public abstract class Try<A> {

    public static boolean isSuccess(Try tryy) {
        return tryy instanceof Success;
    }

    public static <A, B> Function<A, Try<B>> tryOf(ThrowableFunction<A, B> function) {
        return a -> {
            try {
                B result = function.apply(a);
                return new Success<B>(result);
            } catch (Exception e) {
                return new Failure<>(e);
            }
        };
    }

    public abstract boolean isSuccess();

    public boolean isError() {
        return !isSuccess();
    }

    public abstract A getResult();

    public abstract Exception getError();
}

public class Success<A> extends Try<A> {

    private final A result;

    public Success(A result) {
        this.result = result;
    }

    @Override
    public boolean isSuccess() {
        return true;
    }

    @Override
    public A getResult() {
        return result;
    }

    @Override
    public Exception getError() {
        return new UnsupportedOperationException();
    }

    @Override
    public boolean equals(Object that) {
        if(!(that instanceof Success)) {
            return false;
        }
        return Objects.equal(result, ((Success) that).getResult());
    }
}

public class Failure<A> extends Try<A> {

    private final Exception exception;

    public Failure(Exception exception) {
        this.exception = exception;
    }

    @Override
    public boolean isSuccess() {
        return false;
    }

    @Override
    public A getResult() {
        throw new UnsupportedOperationException();
    }

    @Override
    public Exception getError() {
        return exception;
    }
}

一个简单的用例:

List<Try<Integer>> result = Lists.newArrayList(1, 2, 3).stream().
    map(Try.<Integer, Integer>tryOf(i -> someMethodThrowingAnException(i))).
    collect(Collectors.toList());

其他回答

默认情况下,Java 8函数不允许抛出异常,正如在多个回答中所建议的那样,有许多方法来实现它,其中一种方法是:

@FunctionalInterface
public interface FunctionWithException<T, R, E extends Exception> {
    R apply(T t) throws E;
}

定义为:

private FunctionWithException<String, Integer, IOException> myMethod = (str) -> {
    if ("abc".equals(str)) {
        throw new IOException();
    }
  return 1;
};

并在调用者方法中添加抛出或尝试/捕获相同的异常。

免责声明:我还没有使用过Java 8,只是阅读过它。

Function<String, Integer>不抛出IOException,所以你不能在其中放入任何抛出IOException的代码。如果你正在调用一个期望Function<String, Integer>的方法,那么你传递给该方法的lambda不能抛出IOException,句号。你可以这样写一个lambda(我认为这是lambda语法,不确定):

(String s) -> {
    try {
        return myMethod(s);
    } catch (IOException ex) {
        throw new RuntimeException(ex);
        // (Or do something else with it...)
    }
}

或者,如果你传递lambda的方法是你自己写的,你可以定义一个新的函数接口,并使用它作为参数类型,而不是Function<String, Integer>:

public interface FunctionThatThrowsIOException<I, O> {
    O apply(I input) throws IOException;
}

我使用一个名为unchecked()的重载实用程序函数来处理多个用例。


一些用法示例

unchecked(() -> new File("hello.txt").createNewFile());

boolean fileWasCreated = unchecked(() -> new File("hello.txt").createNewFile());

myFiles.forEach(unchecked(file -> new File(file.path).createNewFile()));

支持工具

public class UncheckedUtils {

    @FunctionalInterface
    public interface ThrowingConsumer<T> {
        void accept(T t) throws Exception;
    }

    @FunctionalInterface
    public interface ThrowingSupplier<T> {
        T get() throws Exception;
    }

    @FunctionalInterface
    public interface ThrowingRunnable {
        void run() throws Exception;
    }

    public static <T> Consumer<T> unchecked(
            ThrowingConsumer<T> throwingConsumer
    ) {
        return i -> {
            try {
                throwingConsumer.accept(i);
            } catch (Exception ex) {
                throw new RuntimeException(ex);
            }
        };
    }

    public static <T> T unchecked(
            ThrowingSupplier<T> throwingSupplier
    ) {
        try {
            return throwingSupplier.get();
        } catch (Exception ex) {
            throw new RuntimeException(ex);
        }
    }

    public static void unchecked(
            ThrowingRunnable throwing
    ) {
        try {
            throwing.run();
        } catch (Exception ex) {
            throw new RuntimeException(ex);
        }
    }
}

您需要执行以下操作之一。

If it's your code, then define your own functional interface that declares the checked exception: @FunctionalInterface public interface CheckedFunction<T, R> { R apply(T t) throws IOException; } and use it: void foo (CheckedFunction f) { ... } Otherwise, wrap Integer myMethod(String s) in a method that doesn't declare a checked exception: public Integer myWrappedMethod(String s) { try { return myMethod(s); } catch(IOException e) { throw new UncheckedIOException(e); } } and then: Function<String, Integer> f = (String t) -> myWrappedMethod(t); or: Function<String, Integer> f = (String t) -> { try { return myMethod(t); } catch(IOException e) { throw new UncheckedIOException(e); } };

我会做一些一般的事情:

public interface Lambda {

    @FunctionalInterface
    public interface CheckedFunction<T> {

        T get() throws Exception;
    }

    public static <T> T handle(CheckedFunction<T> supplier) {
        try {
            return supplier.get();
        } catch (Exception exception) {
            throw new RuntimeException(exception);

        }
    }
}

用法:

 Lambda.handle(() -> method());