docs.oracle.com lambda表達式示例將無法編譯

Question

我一直在使用docs.oracle.com作爲學習java的方式，當我試圖編譯下面的代碼示例時，出現了8個錯誤。我正在運行Java 7 u51。好像編譯器沒有認識到oracle教給我的lambda表達式的語法。我真的希望這些教程不會過時，因爲它們是我發現的第一個清楚解釋所有內容的教程。

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); 
    } 
} 

而且錯誤：

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

拉姆達的Java 8.請參閱What's new in Java 8和java.util.function.Consumer@since。