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user1589188
Asked: 2023-10-06 22:02:59 +0800 CST

如何编写抛出相同检查异常的功能接口

  • 772

我知道Runnable不会抛出Exception，所以我尝试编写一个新接口，它接受可以抛出检查异常的方法：

界面：

interface MyInterface {
    void run() throws Exception;
}

服务：

public void serviceMethod(MyInterface action) throws Exception {
    // ...
    action.run();
    // ...
}
public void serviceMethod(Runnable action) { // for methods don't throw checked exception
    // ...
    action.run();
    // ...
}

主要的：

public void myMethod() throws Exception {
    service.serviceMethod(this::someMethodThatThrowsCheckedException);
    service.serviceMethod(this::someMethodThatDoesNotThrowCheckedException);
}

这一切看起来都不错，但我不想myMethod()抛出基类Exception，我希望它抛出实际的异常类someMethodThatThrowsCheckedException。例如如果

public void someMethodThatThrowsCheckedException throws IOException, ClassNotFoundException {//...}

然后我想myMethod()扔IOException, ClassNotFoundException，而不是Exception。就像正常调用方法而不通过接口的情况一样：

public void myMethod() throws IOException, ClassNotFoundException {
    someMethodThatThrowsCheckedException();
}

这可以以通用方式完成吗？

java

1 个回答

  1. Best Answer
    1. 您可以创建抛出通用异常的接口：
        interface MyInterface<T extends Exception> {
        void run() throws T;
    }

    还有一个不会抛出的

        interface MySuperInterface extends MyInterface<Exception> {
        void run();
    }

    用法：

    MyInterface<IOException> io = () -> {throw new IOException();};
MyInterface<IllegalAccessException> iae = () -> {throw new IllegalAccessException();};
    1. 您还可以创建通用消费者方法：

    一种用于投掷：

        public static <T extends Exception> void consume(MyInterface<T> action) throws T {
        action.run();
    }

    一无投掷：

        public static void consume(MySuperInterface action) {
        action.run();
    }
    1. 你可以这样做：
        public static void mainMethod() throws IOException, IllegalAccessException {
        MyInterface<IOException> io = () -> {throw new IOException();};
        MyInterface<IllegalAccessException> iae = () -> {throw new IllegalAccessException();};
        MySuperInterface noException = () -> {};
        //can assign to runnable cause noException.run() doesnt throw an exception
        Runnable runnable = noException::run;

        consume(io);
        consume(iae);
        consume(noException);
    }
    1. 完整示例：
        interface MyInterface<T extends Exception> {
        void run() throws T;
    }

    interface MySuperInterface extends MyInterface<Exception> {
        void run();
    }

    public static void main(String[] args) throws Exception {
        mainMethod();
    }

    public static void mainMethod() throws IOException, IllegalAccessException {
        MyInterface<IOException> io = () -> {throw new IOException();};
        MyInterface<IllegalAccessException> iae = () -> {throw new IllegalAccessException();};
        MySuperInterface noException = () -> {};
        //can assign to runnable cause noException.run() doesnt throw an exception
        Runnable runnable = noException::run;

        consume(io);
        consume(iae);
        consume(noException);
    }

    public static <T extends Exception> void consume(MyInterface<T> action) throws T {
        action.run();
    }

    public static void consume(MySuperInterface action) {
        action.run();
    }

    希望对你有帮助

    • 1

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