Lambda表达式

加上jk。的完美回答，现在你可以使用Lambda表达式(在Java 8中)更容易地传递一个方法。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”:

注意:这使用了一个新的标准函数接口java.util.function.IntConsumer。

class A {
    public static void methodToPass(int i) { 
        // do stuff
    }
}

import java.util.function.IntConsumer;

class B {
    public void dansMethod(int i, IntConsumer aMethod) {
        /* you can now call the passed method by saying aMethod.accept(i), and it
        will be the equivalent of saying A.methodToPass(i) */
    }
}

class C {
    B b = new B();

    public C() {
        b.dansMethod(100, j -> A.methodToPass(j));   //Lambda Expression here
    }
}

上面的例子可以使用::运算符进一步简化。

public C() {
    b.dansMethod(100, A::methodToPass);
}

Java(目前)不支持闭包。但是还有其他语言，如Scala和Groovy，它们运行在JVM中，并且确实支持闭包。

Java很好地支持闭包。它只是不支持函数，所以您习惯的闭包语法更加笨拙和笨重:您必须用一个方法将所有内容包装在一个类中。例如,

public Runnable foo(final int x) {
  return new Runnable() {
    public void run() {
      System.out.println(x);
    }
  };
}

将返回一个Runnable对象，其run()方法“关闭”传入的x，就像在任何支持一级函数和闭包的语言中一样。

我使用的命令模式是@jk。提到，添加一个返回类型:

public interface Callable<I, O> {

    public O call(I input);   
}

Lambda表达式

加上jk。的完美回答，现在你可以使用Lambda表达式(在Java 8中)更容易地传递一个方法。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”:

注意:这使用了一个新的标准函数接口java.util.function.IntConsumer。

class A {
    public static void methodToPass(int i) { 
        // do stuff
    }
}

import java.util.function.IntConsumer;

class B {
    public void dansMethod(int i, IntConsumer aMethod) {
        /* you can now call the passed method by saying aMethod.accept(i), and it
        will be the equivalent of saying A.methodToPass(i) */
    }
}

class C {
    B b = new B();

    public C() {
        b.dansMethod(100, j -> A.methodToPass(j));   //Lambda Expression here
    }
}

上面的例子可以使用::运算符进一步简化。

public C() {
    b.dansMethod(100, A::methodToPass);
}

我知道这是一个相当老的帖子，但我有另一个稍微简单的解决方案。 您可以在其中创建另一个类并使其抽象。接下来，创建一个抽象方法，命名为任何你喜欢的名字。在原来的类中创建一个以新类为参数的方法，在这个方法中调用抽象方法。它看起来是这样的。

public class Demo {

    public Demo(/.../){

    }

    public void view(Action a){
        a.preform();
    }

    /**
     * The Action Class is for making the Demo
     * View Custom Code
     */
    public abstract class Action {

        public Action(/.../){

        }

        abstract void preform();
    }
}

现在您可以执行类似的操作来从类内部调用一个方法。

/...
Demo d = new Demo;
Action a = new Action() {

    @Override
    void preform() {
        //Custom Method Code Goes Here
    }
};

/.../
d.view(a)

就像我说的，我知道它很旧，但我认为这样更容易一些。希望能有所帮助。