How the order of execution is performed between static variables and blocks?

Question

Consider the below scenario:
Code:1

public class StaticDemo {
    static{
        b=5;
        System.out.println("Static B:"+b);/*Compilation error:"Cannot reference a field before it is defined"*/
    }
    static int b;
    static{
        System.out.println("B:"+b);
    }
    public static void main(String[] args) {    
    }
}

Commenting the code as below, there are no errors and following output has been displayed.
Code:2

public class StaticDemo {
    static{
        b=5;
        //System.out.println("Static B:"+b);
    }
    static int b;
    static{
        System.out.println("B:"+b);
    }
    public static void main(String[] args) {    
    }
}

Output-

B:5

If the execution is based on the order in which static variables or blocks have been written.

1. why compilation error is not thrown for the initialization (b=5) as shown in Code:2.

2. And also please explain why error is been thrown for Code:1, if Code:2 is true?

Answer 1

From the Java Language Specification §8.3.2.3

The declaration of a member needs to appear textually before it is used only if the member is an instance (respectively static) field of a class or interface C and all of the following conditions hold:

• The usage occurs in an instance (respectively static) variable initializer of C or in an instance (respectively static) initializer of C.
• The usage is not on the left hand side of an assignment.
• The usage is via a simple name.
• C is the innermost class or interface enclosing the usage.

It is a compile-time error if any of the four requirements above are not met.

[...]

The restrictions above are designed to catch, at compile time, circular or otherwise malformed initializations

In other words, it's OK to write to a field that is declared later in the class, but not to read from it.

There are more detailed examples of this in the example box immediately following the section I have quoted.