Why is this Java code in curly braces ({}) outside of a method?

问题

I am getting ready for a java certification exam and I have seen code LIKE this in one of the practice tests:

class Foo {  
    int x = 1;  
    public static void main(String [] args) {  
        int x = 2;  
        Foo f = new Foo();  
        f.whatever();  
    }  
    { x += x; }  // <-- what's up with this?
    void whatever() {  
        ++x;  
        System.out.println(x);  
    }  
}

My question is ... Is it valid to write code in curly braces outside a method? What are the effects of these (if any)?

回答1:

Borrowed from here -

Normally, you would put code to initialize an instance variable in a constructor. There are two alternatives to using a constructor to initialize instance variables: initializer blocks and final methods. Initializer blocks for instance variables look just like static initializer blocks, but without the static keyword:
{
    // whatever code is needed for initialization goes here
} 
The Java compiler copies initializer blocks into every constructor. Therefore, this approach can be used to share a block of code between multiple constructors.

You may also wanna look at the discussions here.

回答2:

This is an initializer block that is executed while the instance of the class is being loaded/created and that is used to initialize member properties of a class (See Java http://download.oracle.com/javase/tutorial/java/javaOO/initial.html). You can have as many blocks as you want and they will be instantiated from top to bottom.

In addition to the instance block, you can have as many static blocks as you want as well to initialize static members. They would be declared as follows:

public class Initialization {

    static int b = 10;

    int a = 5;

    static {
        b = -9;
    }

    {
        a += 2;
    }

    public static void main(String[] args) throws Exception {

        System.out.println(ClientVoting.b);
        System.out.println(new ClientVoting().a);
        System.out.println(ClientVoting.b);
        System.out.println(new ClientVoting().a);

    }

    static {
        b = 1;
    }

    {
        a++;
    }
}

While the class is being initialized, the static member "b" is initialized as 10, then the first static scope changes its value to -9, and later to 1. This is only executed once while the class is loaded. This executes before the initialization of the first line of the main method.

On the other hand, the similar example to your class is the instance reference "a". A is initialized as 5, then the instance block updates it to 7, and the last block to 8. As expected, the static members are only initialized once in this code, while the instance blocks are executed EVERY time you create a new instance.

The output to this example is 1 8 1 8

回答3:

It's an initializer block. It's used to set instance variables. The motivation to use initializer blocks over constructors is to prevent writing redundant code. The Java compiler copies the contents of the block into each constructor.

来源：https://stackoverflow.com/questions/5865069/why-is-this-java-code-in-curly-braces-outside-of-a-method

标签

java

syntax

braces