java variable scope, variable might not have been initialized

Question

I am studying java, I know Java variable scope, such as class level, method level, block level. However, when I try to practice the variable scope, I meet the error in my code. my code is as following:

public class HelloWorld {
    public static void main(String[] args) {
        int c;
        for (int i=0; i <5; i++) {
            System.out.println(i);
            c = 100;
        }
        System.out.println(c);
    }
}

when I run this code, it display the error: the c variable might not have been initialized, but when I change my code to the following:

public class HelloWorld {
    public static void main(String[] args) {
        int c=0;
        for (int i=0; i <5; i++) {
            System.out.println(i);
            c = 100;
        }
        System.out.println(c);
    }
}

The code will print 100.

How should I understand the scope in my code?

dreamcrash · Accepted Answer · 2020-12-23 10:11:16Z

In Java local variables are not initialized with a by default value (unlike, for example field of classes). From the language specification one (§4.12.5) can read the following:

A local variable (§14.4, §14.14) must be explicitly given a value before it is used, by either initialization (§14.4) or assignment (§15.26), in a way that can be verified using the rules for definite assignment (§16 (Definite Assignment)).

Because it is explicitly set on the Java language specification the compiler will not try to (and should not) infer that the variable c will always be updated inside the loop:

public class HelloWorld {
    public static void main(String[] args) {
        int c;
        for (int i=0; i <5; i++) {
            System.out.println(i);
            c = 100;
        }
        System.out.println(c);
    }
}

The compiler strictly enforces the standard and notifies you about having breaking one of its rules with the error:

"variable c might not have been initialized"

So even though your code can be formally proven to be valid, it is not the compiler job to try to analyze your application's logic, and neither does the rules of local variable initialization rely on that. The compiler checks if the variable c is initialized according to the local variable initialization rules, and reacts accordingly (e.g., displaying a compilation error for the case of int c;).

Vitaly Kolesnikov · Accepted Answer · 2020-12-23 09:17:31Z

0

This is not a scope issue. Your variable c is method-scoped and available within whole main method. It's a compiler feature. At compile time it can't check that "for" cycle will run at least once to initialize the variable. That`s why it says that variable might not have been initialized when you are trying to print it. So c is available to print method, but compiler is not sure it was initialized.

answered Dec 23, 2020 at 9:17

Vitaly Kolesnikov

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Thuat Nguyen · Accepted Answer · 2020-12-23 09:18:23Z

0

Because you can't make sure variable c is always initialized in the loop. For example if your code look like this:

public class HelloWorld {
    public static void main(String[] args) {
        int c;
        for (int i=0; i <0; i++) {
            System.out.println(i);
            c = 100;
        }
        System.out.println(c);
    }
}

Your code will never reach the loop then variable c never is initialized. That's why java language request you initialize a variable outside a loop if you use it outside a loop.

answered Dec 23, 2020 at 9:18

Thuat Nguyen

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