Question

C# lets me make arrays on the fly when I need to pass them into functions. Let’s say I have a method called findMiddleItem(String[] items). In C#, I can write code like:

findMiddleItem(new String[] { "one", "two", "three" });

It’s awesome, because it means I don’t have to write:

IList<String> strings = new List<String>();
strings.add("one");
strings.add("two");
strings.add("three");
findMiddleItem(strings.ToArray());

Which sucks, because I don’t really care about strings — it’s just a construct to let me pass a string array into a method that requires it. A method which I can’t modify.

So how do you do this in Java? I need to know this for array types (eg. String[]) but also generic types (eg. List).

Answer 1

A List and an Array are fundamentally different things.

A List is a Collection type, an implementation of an interface.
An Array is a special operating system specific data structure that can only be created through either a special syntax or native code.

Arrays

In Java, the array syntax is identical to the one you are describing:

String[] array = new String[] { "one", "two", "three" };

Reference: Java tutorial > Arrays

Lists

The easiest way to create a List is this:

List<String> list = Arrays.asList("one", "two", "three");

However, the resulting list will be immutable (or at least it won’t support add() or remove()), so you can wrap the call with an ArrayList constructor call:

new ArrayList<String>(Arrays.asList("one", "two", "three"));

As Jon Skeet says, it’s prettier with Guava, there you can do:

Lists.newArrayList("one", "two", "three");

Reference: Java Tutorial > The List Interface, Lists (guava javadocs)

VarArgs

About this comment:

It would be nice if we would be able to do findMiddleItem({ "one", "two", "three" });

Java varargs gives you an even better deal:

public void findMiddleItem(String ... args){
    //
}

you can call this using a variable number of arguments:

findMiddleItem("one");
findMiddleItem("one", "two");
findMiddleItem("one", "two", "three");

Or with an array:

findMiddleItem(new String[]{"one", "two", "three"});

Reference: Java Tutorial > Arbitrary Number of Arguments