Using Generics on right hand side in Java 6?

问题

I java 6 i can declare the arraylist as follows

Way1: using generics i.e <Integer> on right hand side too

List<Integer> p = new ArrayList<Integer>();

Way2: using the diamond operator

List<Integer> p = new ArrayList<>();

Way3: using generic only at left side

List<Integer> p = new ArrayList();

I prefer to use way 3 as its brief. Is there any difference between these ways? Which one we should prefer and why?

Update:- I know in java 7 second way is recommended but my question is in context of java 6. Which one is preferable?

To me, way 3 also says p is an arraylist of integers (same conveyed by way1). So I find no difference except the fact IDE displays warning message:

ArrayList is a raw type. References to generic type ArrayList<E> should be parameterized

回答1:

As has been pointed out, Way 2 isn't valid in 1.6. So the question is, is there any difference between Way 1 and Way 3. Apart from readability, no.

Take this code:

import java.util.*;
class G {
  public static void main(String [] args){
    List<Integer> listOne = new ArrayList<Integer>();
    listOne.add(1);
    int one = listOne.get(0);

    List<Integer> listTwo = new ArrayList();
    listTwo.add(1);
    int two = listTwo.get(0);
  }
}

Compile it and look at the bytecode using javap -c

  public static void main(java.lang.String[]);
    Code:
    // FOR listOne
       0: new           #2                  // class java/util/ArrayList
       3: dup           
       4: invokespecial #3                  // Method java/util/ArrayList."<init>":()V
       7: astore_1      
       8: aload_1       
       9: iconst_1      
      10: invokestatic  #4                  // Method java/lang/Integer.valueOf:(I)Ljava/lang/Integer;
      13: invokeinterface #5,  2            // InterfaceMethod java/util/List.add:(Ljava/lang/Object;)Z
      18: pop           
      19: aload_1       
      20: iconst_0      
      21: invokeinterface #6,  2            // InterfaceMethod java/util/List.get:(I)Ljava/lang/Object;
      26: checkcast     #7                  // class java/lang/Integer
      29: invokevirtual #8                  // Method java/lang/Integer.intValue:()I
      32: istore_2      
   // FOR listTwo
      33: new           #2                  // class java/util/ArrayList
      36: dup           
      37: invokespecial #3                  // Method java/util/ArrayList."<init>":()V
      40: astore_3      
      41: aload_3       
      42: iconst_1      
      43: invokestatic  #4                  // Method java/lang/Integer.valueOf:(I)Ljava/lang/Integer;
      46: invokeinterface #5,  2            // InterfaceMethod java/util/List.add:(Ljava/lang/Object;)Z
      51: pop           
      52: aload_3       
      53: iconst_0      
      54: invokeinterface #6,  2            // InterfaceMethod java/util/List.get:(I)Ljava/lang/Object;
      59: checkcast     #7                  // class java/lang/Integer
      62: invokevirtual #8                  // Method java/lang/Integer.intValue:()I
      65: istore        4
      67: return        
}

We can see that the exact same bytecode is produced in both cases. Note that as Generics aren't baked in the compiler throws away the information after checking it at compile time and adds in checkcast instructions to make sure the casts it does when retrieving objects are safe.

回答2:

The second way is not possible in Java 6. It is the new way of inferring Generic instance in Java 7.

回答3:

There is no difference, if you are using java 7 prefer the second method, but is it not available in java 6. It is a new addition to java 7.

回答4:

Both are same but way2 is available from java 7

回答5:

Both are same .But there are version difference in both of this.

But second way is not possible in java 6.In Java 7 if we not declare type in right side then by default it will take same type as left side.

http://docs.oracle.com/javase/7/docs/technotes/guides/language/type-inference-generic-instance-creation.html

So as per your updates, if you have to use Java 6,then you should use way1.

回答6:

Way 3 uses raw types. You should never use raw types. They should only be used in legacy code.

回答7:

Way 3 is not good. Mixing Generics and raw types is naughty, as you are making an assumption for runtime about types, and can run into ClassCastExceptions like the following code:

ArrayList b = new ArrayList();
b.add(5);
ArrayList<String> a = new ArrayList(b);
System.out.println(a.get(0));

So for Java 6, always use way 1

来源：https://stackoverflow.com/questions/17357883/using-generics-on-right-hand-side-in-java-6

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