docs.oracle.com lambda表达式示例将无法编译

Question

我一直在使用docs.oracle.com作为学习java的方式，当我试图编译下面的代码示例时，出现了8个错误。我正在运行Java 7 u51。好像编译器没有认识到oracle教给我的lambda表达式的语法。我真的希望这些教程不会过时，因为它们是我发现的第一个清楚解释所有内容的教程。

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); 
    } 
} 

而且错误：

C:\java>javac LambdaScopeTest.java 
LambdaScopeTest.java:19: illegal start of expression 
      Consumer<Integer> myConsumer = (y) -> 
               ^
LambdaScopeTest.java:20: illegal start of expression 
      { 
      ^
LambdaScopeTest.java:28: <identifier> expected 
      myConsumer.accept(x); 
          ^
LambdaScopeTest.java:28: <identifier> expected 
      myConsumer.accept(x); 
          ^
LambdaScopeTest.java:33: class, interface, or enum expected 
    public static void main(String... args) { 
       ^
LambdaScopeTest.java:35: class, interface, or enum expected 
     LambdaScopeTest.FirstLevel fl = st.new FirstLevel(); 
     ^
LambdaScopeTest.java:36: class, interface, or enum expected 
     fl.methodInFirstLevel(23); 
     ^
LambdaScopeTest.java:37: class, interface, or enum expected 
    } 
    ^
8 errors 

Answer 1

拉姆达的Java 8.请参阅What's new in Java 8和java.util.function.Consumer@since。