我正在寻找一种通过引用传递方法的方法。我知道Java不传递方法作为参数,但是,我想要一个替代方案。

我被告知接口是作为参数传递方法的替代方案,但我不理解接口如何通过引用充当方法。如果我理解正确的话,接口只是一组没有定义的抽象方法。我不想发送一个每次都需要定义的接口,因为几个不同的方法可以用相同的参数调用相同的方法。

我想要完成的是类似这样的事情:

public void setAllComponents(Component[] myComponentArray, Method myMethod) {
    for (Component leaf : myComponentArray) {
        if (leaf instanceof Container) { //recursive call if Container
            Container node = (Container) leaf;
            setAllComponents(node.getComponents(), myMethod);
        } //end if node
        myMethod(leaf);
    } //end looping through components
}

调用方法如下:

setAllComponents(this.getComponents(), changeColor());
setAllComponents(this.getComponents(), changeSize());

当前回答

带有反射的解的例子,通过的方法必须是公共的

import java.lang.reflect.Method;
import java.lang.reflect.InvocationTargetException;

public class Program {
    int i;

    public static void main(String[] args) {
        Program   obj = new Program();    //some object

        try {
            Method method = obj.getClass().getMethod("target");
            repeatMethod( 5, obj, method );
        } 
        catch ( NoSuchMethodException | IllegalAccessException | InvocationTargetException e) {
            System.out.println( e ); 
        }
    }

    static void repeatMethod (int times, Object object, Method method)
        throws IllegalAccessException, InvocationTargetException {

        for (int i=0; i<times; i++)
            method.invoke(object);
    }
    public void target() {                 //public is necessary
        System.out.println("target(): "+ ++i);
    }
}

其他回答

首先用要作为参数传递的方法定义一个Interface

public interface Callable {
  public void call(int param);
}

使用该方法实现一个类

class Test implements Callable {
  public void call(int param) {
    System.out.println( param );
  }
}

//像这样调用

Callable cmd = new Test();

这允许您将cmd作为参数传递,并调用接口中定义的方法调用

public invoke( Callable callable ) {
  callable.call( 5 );
}

java 8起

Java 8以后,您可以使用lambda表达式提供函数接口(只有一个抽象方法的接口)的抽象方法的实现,并将其作为参数传递给方法。

@FunctionalInterface
interface ArithmeticFunction {
    public int calcualate(int a, int b);
}

public class Main {
    public static void main(String args[]) {
        ArithmeticFunction addition = (a, b) -> a + b;
        ArithmeticFunction subtraction = (a, b) -> a - b;

        int a = 20, b = 5;

        System.out.println(perform(addition, a, b));
        // or
        System.out.println(perform((x, y) -> x + y, a, b));

        System.out.println(perform(subtraction, a, b));
        // or
        System.out.println(perform((x, y) -> x - y, a, b));
    }

    static int perform(ArithmeticFunction function, int a, int b) {
        return function.calcualate(a, b);
    }
}

输出:

25
25
15
15

在线演示

从方法参考中了解更多信息。

我不是java专家,但我可以这样解决你的问题:

@FunctionalInterface
public interface AutoCompleteCallable<T> {
  String call(T model) throws Exception;
}

我在我的特殊接口中定义了参数

public <T> void initialize(List<T> entries, AutoCompleteCallable getSearchText) {.......
//call here
String value = getSearchText.call(item);
...
}

最后,我实现getSearchText方法,同时调用初始化方法。

initialize(getMessageContactModelList(), new AutoCompleteCallable() {
          @Override
          public String call(Object model) throws Exception {
            return "custom string" + ((xxxModel)model.getTitle());
          }
        })

我很欣赏上面的答案,但我可以使用下面的方法实现相同的行为;这个想法借鉴自Javascript的回调。我对修正持开放态度,尽管到目前为止(在生产中)还不错。

其思想是在签名中使用函数的返回类型,这意味着yield必须是静态的。

下面是一个运行带有超时的进程的函数。

public static void timeoutFunction(String fnReturnVal) {

    Object p = null; // whatever object you need here

    String threadSleeptime = null;

    Config config;

    try {
        config = ConfigReader.getConfigProperties();
        threadSleeptime = config.getThreadSleepTime();

    } catch (Exception e) {
        log.error(e);
        log.error("");
        log.error("Defaulting thread sleep time to 105000 miliseconds.");
        log.error("");
        threadSleeptime = "100000";
    }

    ExecutorService executor = Executors.newCachedThreadPool();
    Callable<Object> task = new Callable<Object>() {
        public Object call() {
            // Do job here using --- fnReturnVal --- and return appropriate value
            return null;
        }
    };
    Future<Object> future = executor.submit(task);

    try {
        p = future.get(Integer.parseInt(threadSleeptime), TimeUnit.MILLISECONDS);
    } catch (Exception e) {
        log.error(e + ". The function timed out after [" + threadSleeptime
                + "] miliseconds before a response was received.");
    } finally {
        // if task has started then don't stop it
        future.cancel(false);
    }
}

private static String returnString() {
    return "hello";
}

public static void main(String[] args) {
    timeoutFunction(returnString());
}

在Java 8中,您现在可以使用Lambda表达式和方法引用更容易地传递方法。首先介绍一些背景知识:函数式接口是具有且仅有一个抽象方法的接口,尽管它可以包含任意数量的默认方法(Java 8新增)和静态方法。lambda表达式可以快速实现抽象方法,如果不使用lambda表达式,则无需使用所有不必要的语法。

没有lambda表达式:

obj.aMethod(new AFunctionalInterface() {
    @Override
    public boolean anotherMethod(int i)
    {
        return i == 982
    }
});

使用lambda表达式:

obj.aMethod(i -> i == 982);

以下是Java Lambda表达式教程的节选:

Syntax of Lambda Expressions A lambda expression consists of the following: A comma-separated list of formal parameters enclosed in parentheses. The CheckPerson.test method contains one parameter, p, which represents an instance of the Person class.Note: You can omit the data type of the parameters in a lambda expression. In addition, you can omit the parentheses if there is only one parameter. For example, the following lambda expression is also valid: p -> p.getGender() == Person.Sex.MALE && p.getAge() >= 18 && p.getAge() <= 25 The arrow token, -> A body, which consists of a single expression or a statement block. This example uses the following expression: p.getGender() == Person.Sex.MALE && p.getAge() >= 18 && p.getAge() <= 25 If you specify a single expression, then the Java runtime evaluates the expression and then returns its value. Alternatively, you can use a return statement: p -> { return p.getGender() == Person.Sex.MALE && p.getAge() >= 18 && p.getAge() <= 25; } A return statement is not an expression; in a lambda expression, you must enclose statements in braces ({}). However, you do not have to enclose a void method invocation in braces. For example, the following is a valid lambda expression: email -> System.out.println(email) Note that a lambda expression looks a lot like a method declaration; you can consider lambda expressions as anonymous methods—methods without a name.


下面是如何使用lambda表达式“传递一个方法”:

interface I {
    public void myMethod(Component component);
}

class A {
    public void changeColor(Component component) {
        // code here
    }

    public void changeSize(Component component) {
        // code here
    }
}
class B {
    public void setAllComponents(Component[] myComponentArray, I myMethodsInterface) {
        for(Component leaf : myComponentArray) {
            if(leaf instanceof Container) { // recursive call if Container
                Container node = (Container)leaf;
                setAllComponents(node.getComponents(), myMethodInterface);
            } // end if node
            myMethodsInterface.myMethod(leaf);
        } // end looping through components
    }
}
class C {
    A a = new A();
    B b = new B();

    public C() {
        b.setAllComponents(this.getComponents(), component -> a.changeColor(component));
        b.setAllComponents(this.getComponents(), component -> a.changeSize(component));
    }
}

类C可以通过使用类似这样的方法引用来进一步缩短:

class C {
    A a = new A();
    B b = new B();

    public C() {
        b.setAllComponents(this.getComponents(), a::changeColor);
        b.setAllComponents(this.getComponents(), a::changeSize);
    }
}