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Why illegal forward reference error not shown while using static variable with class name



In the below code while accessing a static variable with class name it doesn't throw a forward reference error but accessing it without class name does.

Why this doesn't happen when accessing with class name?

class Test{
    static {
        System.out.println(a); // shows error
        a = 99; // and this line too doesn't give error  
        System.out.println(Test.a); // this line doesn't
    static int a = 10;  
like image 317
Prasanna Avatar asked Mar 17 '23 23:03


1 Answers

The rules for forward reference is defined in JLS §8.3.3:

Use of class variables whose declarations appear textually after the use is sometimes restricted, even though these class variables are in scope (§6.3). Specifically, it is a compile-time error if all of the following are true:

  • The declaration of a class variable in a class or interface C appears textually after a use of the class variable;

  • The use is a simple name in either a class variable initializer of C or a static initializer of C;

  • The use is not on the left hand side of an assignment;

  • C is the innermost class or interface enclosing the use.

So, basically your first Sysout(), satisfies all the above 4 conditions, and hence it's a compile time error.

In the 2nd Sysout(), you're accessing a using it's qualified name, rather than simple name, which as per the above rules is allowed.

Now, the reason for this would be, when you access Test.a, the compiler is sure that Test class has been loaded and all static fields have been initialized, so it can access the field a. But while accessing a on simple name, the compiler isn't sure that initializer for a has already run or not, since it might still be in process of loading the class.

Consider the following process of loading of class:

  • When a class is loaded, memory is allocated for all the static variables declared in it. At this point, the variable a has got its memory allocated (declaration done)
  • Then all the static initializers run in the order of occurrence.
    • First statement is Sysout(a);. a hasn't been initialized yet, so you can't access it. (error)
    • Second statement is a = 99. Here you're actually initializing the variable a. Perfectly fine.
    • Third one is Sysout(Test.a) - reasoning for this is already posted above. Compiler knows Test is already loaded.
    • Then static int a = 10 is executed. It re-initializes a to 10. Remember, declaration part is already taken care of in first step.
like image 155
Rohit Jain Avatar answered Mar 25 '23 04:03

Rohit Jain