java8lambda表达式

Java 8 Lambda 表达式
语法
lambda 表达式的语法格式如下：

(parameters) -> expression
或
(parameters) ->{ statements; }

Lambda 表达式实例

public class Calculator {

    interface IntegerMath {
        int operation(int a, int b);
    }

    public int operateBinary(int a, int b, IntegerMath op) {
        return op.operation(a, b);
    }

    public static void main(String... args) {
        Calculator myApp = new Calculator();
        IntegerMath addition = (a, b) -> a + b;
        IntegerMath subtraction = (a, b) -> a - b;
        System.out.println("40 + 2 = " + myApp.operateBinary(40, 2, addition));
        System.out.println("20 - 10 = " + myApp.operateBinary(20, 10, subtraction));
    }
}

点击查看结果

public class Java8Lambda {
    public static void main(String args[]) {
        Java8Lambda tester = new Java8Lambda();

        // 类型声明
        MathOperation addition = (int a, int b) -> a + b;

        // 不用类型声明
        MathOperation subtraction = (a, b) -> a - b;

        // 大括号中的返回语句
        MathOperation multiplication = (int a, int b) -> {
            return a * b;
        };

        // 没有大括号及返回语句
        MathOperation division = (int a, int b) -> a / b;

        System.out.println("10 + 5 = " + tester.operate(10, 5, addition));
        System.out.println("10 - 5 = " + tester.operate(10, 5, subtraction));
        System.out.println("10 x 5 = " + tester.operate(10, 5, multiplication));
        System.out.println("10 / 5 = " + tester.operate(10, 5, division));

        // 不用括号
        GreetingService greetService1 = message -> System.out.println("Hello " + message);

        // 用括号
        GreetingService greetService2 = (message) -> System.out.println("Hello " + message);

        greetService1.sayMessage("Runoob");
        greetService2.sayMessage("Google");
    }

    interface MathOperation {
        int operation(int a, int b);
    }

    interface GreetingService {
        void sayMessage(String message);
    }

    private int operate(int a, int b, MathOperation mathOperation) {
        return mathOperation.operation(a, b);
    }
}

点击查看结果

变量作用域
lambda 表达式只能引用标记了 final 的外层局部变量，这就是说不能在 lambda 内部修改定义在域外的局部变量，否则会编译错误。

import java.util.function.Consumer;

public class LambdaScopeTest {

    public int x = 0;

    class FirstLevel {
        public int x = 1;

        void methodInFirstLevel(int x) {

            // The following statement causes the compiler to generate
            // the error "local variables referenced from a lambda expression
            // must be final or effectively final" in statement A:
            //
            // x = 99;

            Consumer<Integer> myConsumer = (y) -> {
                System.out.println("x = " + x); // Statement A
                System.out.println("y = " + y);
                System.out.println("this.x = " + this.x);
                System.out.println("LambdaScopeTest.this.x = " + LambdaScopeTest.this.x);
            };

            myConsumer.accept(x);
        }
    }

    public static void main(String... args) {
        LambdaScopeTest st = new LambdaScopeTest();
        LambdaScopeTest.FirstLevel fl = st.new FirstLevel();
        fl.methodInFirstLevel(23);
    }
}

点击查看结果

lambda 表达式的局部变量可以不用声明为 final，但是必须不可被后面的代码修改。（即隐性的具有 final 的语义）

public class Java8Tester {
    public static void main(String args[]) {
        final int num = 1;
        Converter<Integer, String> s = (param) -> System.out.println(String.valueOf(param + num));
        s.convert(2);  // 输出结果为 3
    }
 
    public interface Converter<T1, T2> {
        void convert(int i);
    }
}

在 Lambda 表达式当中不允许声明一个与局部变量同名的参数或者局部变量。

参考资料

• What's New in JDK 8
• Java 8 新特性
• http://www.cnblogs.com/onetwo/p/8526374.html
• Lambda Expressions
• https://blog.csdn.net/yitian_66/article/details/81010434