1.lambada的存在来由

匿名类的一个问题是，如果匿名类的实现非常简单，例如只包含一个方法的接口，那么匿名类的语法可能看起来不实用且不清楚。在这些情况下，您通常会尝试将功能作为参数传递给另一个方法，例如当有人单击按钮时应采取的操作。Lambda表达式使您可以执行此操作，将功能视为方法参数，或将代码视为数据。

单 interface 单method

public interface Predicate<T> {

    /**     * Evaluates this predicate on the given argument.     *     * @param t the input argument     * @return {@code true} if the input argument matches the predicate,     * otherwise {@code false}     */    boolean test(T t);

2.lambada理想用例

参考 java.util.function 下面的个别functionnal interfaces

Consequently, the JDK defines several standard functional interfaces, which you can find in the package java.util.function.

定义一个 Person 类

public class Person {public enum Sex {MALE, FEMALE}String name;LocalDate birthday;Sex gender;String emailAddress;public int getAge() {// ...}public void printPerson() {// ...}
}

有一个List<Person> 我们要打印出指定条件的Person信息，用lamdaba表达式实现可以这么做

方式一

public static void listConditionPerson(List<Person> person, Predicate<Person> pre, Consumer<Person> conf) {    for (Person p : person) {        if (pre.test(p))            conf.accept(p);    }}

调用方式为：　　

listConditionPerson(person, person1 -> person1.getAge() > 10 && person1.getAge() <= 30, Person::printPerson);

方式二：使用泛型（generics）的模式改造一下方法，可以支持不同队形的相同功能

public static <X, Y> void processElements(Iterable<X> source,Predicate<X> tester,Function <X, Y> mapper,Consumer<Y> block) {for (X p : source) {if (tester.test(p)) {Y data = mapper.apply(p);block.accept(data);}}
}

调用方式为

 processElements(        person,        p -> p.getGender() == Person.Sex.MALE                && p.getAge() >= 18                && p.getAge() <= 25,        p -> p.getEmailAddress(),        email -> System.out.println(email));

方式三:使用支持lambada的聚合函数（stream api）

person.stream().filter(person1 -> person1.getAge() > 10).map(p ->    p.getEmailAddress()).forEach(email -> {    System.out.println(email);});

并发版

person.parallelStream().filter(person1 -> person1.getAge() > 10).map(p ->        p.getEmailAddress()).forEach(email -> {    System.out.println(email);});

lambda 语法官方介绍

bda expression looks a lot like a method declaration; you can consider lambda expressions as anonymous methods—methods without a name.

The following example, Calculator, is an example of lambda expressions that take more than one formal parameter:

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));    }
}

The method operateBinary performs a mathematical operation on two integer operands. The operation itself is specified by an instance of IntegerMath. The example defines two operations with lambda expressions, addition and subtraction. The example prints the following:

40 + 2 = 42
20 - 10 = 10

Accessing Local Variables of the Enclosing Scope

Like local and anonymous classes, lambda expressions can capture variables; they have the same access to local variables of the enclosing scope. However, unlike local and anonymous classes, lambda expressions do not have any shadowing issues (see Shadowing for more information). Lambda expressions are lexically scoped. This means that they do not inherit any names from a supertype or introduce a new level of scoping. Declarations in a lambda expression are interpreted just as they are in the enclosing environment. The following example, LambdaScopeTest, demonstrates this:

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 ASystem.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);}
}

This example generates the following output:

x = 23
y = 23
this.x = 1
LambdaScopeTest.this.x = 0

If you substitute the parameter x in place of y in the declaration of the lambda expression myConsumer, then the compiler generates an error:

Consumer<Integer> myConsumer = (x) -> {// ...
}

The compiler generates the error "variable x is already defined in method methodInFirstLevel(int)" because the lambda expression does not introduce a new level of scoping. Consequently, you can directly access fields, methods, and local variables of the enclosing scope. For example, the lambda expression directly accesses the parameter x of the method methodInFirstLevel. To access variables in the enclosing class, use the keyword this. In this example, this.x refers to the member variable FirstLevel.x.

However, like local and anonymous classes, a lambda expression can only access local variables and parameters of the enclosing block that are final or effectively final. For example, suppose that you add the following assignment statement immediately after the methodInFirstLevel definition statement:

void methodInFirstLevel(int x) {x = 99;// ...
}

Because of this assignment statement, the variable FirstLevel.x is not effectively final anymore. As a result, the Java compiler generates an error message similar to "local variables referenced from a lambda expression must be final or effectively final" where the lambda expression myConsumer tries to access the FirstLevel.x variable:

System.out.println("x = " + x);

Target Typing

How do you determine the type of a lambda expression? Recall the lambda expression that selected members who are male and between the ages 18 and 25 years:

p -> p.getGender() == Person.Sex.MALE&& p.getAge() >= 18&& p.getAge() <= 25

This lambda expression was used in the following two methods:

• public static void printPersons(List<Person> roster, CheckPerson tester) in Approach 3: Specify Search Criteria Code in a Local Class

• public void printPersonsWithPredicate(List<Person> roster, Predicate<Person> tester) in Approach 6: Use Standard Functional Interfaces with Lambda Expressions

When the Java runtime invokes the method printPersons, it's expecting a data type of CheckPerson, so the lambda expression is of this type. However, when the Java runtime invokes the method printPersonsWithPredicate, it's expecting a data type of Predicate<Person>, so the lambda expression is of this type. The data type that these methods expect is called the target type. To determine the type of a lambda expression, the Java compiler uses the target type of the context or situation in which the lambda expression was found. It follows that you can only use lambda expressions in situations in which the Java compiler can determine a target type:

• Variable declarations

• Assignments

• Return statements

• Array initializers

• Method or constructor arguments

• Lambda expression bodies

• Conditional expressions, ?:

• Cast expressions

Target Types and Method Arguments

For method arguments, the Java compiler determines the target type with two other language features: overload resolution and type argument inference.

Consider the following two functional interfaces ( java.lang.Runnable and java.util.concurrent.Callable<V>):

public interface Runnable {void run();
}public interface Callable<V> {V call();
}

The method Runnable.run does not return a value, whereas Callable<V>.call does.

Suppose that you have overloaded the method invoke as follows (see Defining Methods for more information about overloading methods):

void invoke(Runnable r) {r.run();
}<T> T invoke(Callable<T> c) {return c.call();
}

Which method will be invoked in the following statement?

String s = invoke(() -> "done");

The method invoke(Callable<T>) will be invoked because that method returns a value; the method invoke(Runnable) does not. In this case, the type of the lambda expression () -> "done" is Callable<T>.

博客新手，勿喷！