Java中如何将函数名作为参数传递

采用lambda表达式：

import java.util.function.Function;
class Test {

	public static void main(String... args) {
		Function<Integer, Integer> increase = e -> e + 7;  // lambda表达式

		System.out.println(increase.getClass());

		funcPlus(3, increase);

	}

	public static void funcPlus(int value, Function<Integer, Integer> func) {
		System.out.println(func.apply(value));
	}

}

输出结果：

class com.classTest.Test$$Lambda$1/0x0000000801200840
10

（1）apply() 函数在最新的1.8 Java版本才支持 

java.util.function
Interface Function<T,R>
Type Parameters:
T - the type of the input to the function
R - the type of the result of the function
All Known Subinterfaces:
UnaryOperator<T>
Functional Interface:
This is a functional interface and can therefore be used as the assignment target for a lambda expression or method reference.

@FunctionalInterface
public interface Function<T,R>
Represents a function that accepts one argument and produces a result.
This is a functional interface whose functional method is apply(Object).

Since:
1.8

 API文档：https://docs.oracle.com/javase/8/docs/api/java/util/function/Function.html

采用Callable方式：

import java.util.concurrent.Callable;

public class CallableUse {
	public static void main(String... args) {
		
//		final int num = 100770;  or
		int num = 100770;
		// 使用匿名的内部类, 如果需要传递参数可能需要将变量转换成final：
		try {
			callMethod(100, new Callable<Integer>() {
				public Integer call() {
					return needOperation(num);
				}
			});
		} catch (Exception e1) {
			// TODO Auto-generated catch block
			e1.printStackTrace();
		}
		
	}


	public static int needOperation(int param) {
		// do something
		param = 999;
		return param;
	}

	public static void callMethod(int i, Callable<Integer> myFunc)  {
		// do something
		try {
			System.out.println(myFunc.call() );
		} catch (Exception e) {
			// TODO Auto-generated catch block
			e.printStackTrace();
		}
		
	}

}

输出结果：

999

 附：

同时在这篇帖子上有人采用Java反射机制：

https://stackoverflow.com/questions/4685563/how-to-pass-a-function-as-a-parameter-in-java

import java.lang.reflect.Method;

public class Demo {

    public static void main(String[] args) throws Exception{
        Class[] parameterTypes = new Class[1];
        parameterTypes[0] = String.class;
        Method method1 = Demo.class.getMethod("method1", parameterTypes);

        Demo demo = new Demo();
        demo.method2(demo, method1, "Hello World");
    }

    public void method1(String message) {
        System.out.println(message);
    }

    public void method2(Object object, Method method, String message) throws Exception {
        Object[] parameters = new Object[1];
        parameters[0] = message;
        method.invoke(object, parameters);
    }

}

  

参考资料：

https://techndeck.com/how-to-pass-function-as-a-parameter-in-a-method-in-java-8/