How Java initializes ArrayList with a single line of code

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I would like to create a list of options to test something.

I am doing:

  arraylist<string> places = new  arraylist<string > ();  
 Places. "Buenos Aires"  Places. Add Places. Add "La Plata"                  /span>                

I refactor the code to do:

ArrayList<String> places = new ArrayList<String>(Arrays.asList("Buenos Aires", "Córdoba", "La Plata"));

Is there a better way to do that?

Thanks for reading!

Workaround 1:

In fact, you might want to initialize the "best" way, ArrayList is the way you write, because it doesn't need to create a new List in any way:

 arraylist<string>  list = newarraylist<string> ();      List. Add "A"      List. Add "B"      List. Add "C"                  /span>          

Catch is quite a lot of typing required, see list instances.

A method that resembles an instance initializer (also known as a "double curly brace Initialization") for an anonymous inner class:

  arraylist<string> list = new  arraylist<string > ()  {{ 
     Add (  "A" );  
     Add ( "B"      Add ( "C"  }} 
           /span>                

However, I don't really like the method because what you end up being a subclass of a class ArrayList has an instance initializer and this class just to create an object--just looks like I'm a bit overqualified.

What's good is if you accept the item coin Collection text Suggestions, we can have the text in the list in Java 7:

List<String> list = ["A", "B", "C"];

Unfortunately it will not help you here, as it will initialize one List , and not ArrayList , moreover, it has not yet been provided if it ever will and.

Workaround 2:

Would it be easier if you just declared it to be ArrayList in the list?

List<String> places = Arrays.asList("Buenos Aires", "Córdoba", "La Plata");

Workaround 3:

ImportCOM.  Google.  Common.  Collect.  Immutablelist;

....

List<String> places = immutablelist.  of ("Buenos Aires", "Córdoba", "La Plata");

Workaround 4:

Pretty much, the simplest way is List<String> strings = new ArrayList<String>(asList("foo", "bar", "baz")) if you have to have one ArrayList , otherwise you should just use it in most casesList<String> strings = asList("foo", "bar", "baz")

You say you have announced as a list one ArrayList in your code, but you should not do it unless you are using some members ArrayList in this is not List , for example ArrayList.ensureCapacity() .

Usually you should just declare variables that you want to use, and its list is usually the most common interface List , and initialize them in concrete implementations, for example ArrayList or LinkedList .

Use is not implemented, otherwise you will find an interface that you want to change in multiple locations when you want to use another implementation.

Another example will always declare variables, one InputStream even if it is usually one FileInputStream , which is because one day soon you or someone else will want to use some alternative InputStream .

Workaround 5:

You can create a factory method:

Public Static ArrayList<String>Createarraylist(String ...Elements) {
ArrayList<String>List= New ArrayList<String> (); 
For (StringElement:Elements) {
List.Addelement   } 
   return List 
 } 
 
  ....  
 
 arraylist<string > places = createarraylist  ( 
    "S?o Paulo"   "Rio de Janeiro" ,  Brasília ");  
               /span>                

But it's no more than the first time you refactor.

Update: greater flexibility, it can be a common:

Public Static <T> ArrayList<T>Createarraylist(T...Elements) { 
ArrayList<T>List= new arraylist<t> ();    
   for  (t element : Elements)  { 
     List. Addelement   } 
   return List 
 } 
         /span>                

Workaround 6:

Just notice it in a very simple working method as follows:

ArrayListArrlist= New ArrayList() {"1",2,3,"4" }; 
 
 list<customer > Listcustomer = new  list<customer> ()  { new  Customer (),  new  Customer (),  new  Customer ()  };                 /span>                

This C # 3.0 does not support the work required by the two. Hope this helps.

Workaround 7:

To set a copy of the default object for the list to populate N:

ArrayList<Object> list = new ArrayList<Object>(
    Collections.nCopies(1000, new Object()));

How Java initializes ArrayList with a single line of code