Java: Why am I required to initialize a primitive local variable?

user1329572 picture user1329572 · Jun 23, 2012 · Viewed 13.1k times · Source
public class Foo {
    public static void main(String[] args) {
        float f;
        System.out.println(f);
    }
}

The print statement causes the following compile-time error,

The local variable f may not have been initialized

If primitives in Java already have a default value (float = 0.0f), why am I required to define one?


Edit:

So, this works

public class Foo {
    float f;
    public static void main(String[] args) {
        System.out.println(new Foo().f);
    }
}

Thanks, everyone!

Answer

Zakaria picture Zakaria · Jun 23, 2012

Because it's a local variable. This is why nothing is assigned to it :

Local variables are slightly different; the compiler never assigns a default value to an uninitialized local variable. If you cannot initialize your local variable where it is declared, make sure to assign it a value before you attempt to use it. Accessing an uninitialized local variable will result in a compile-time error.

Edit: Why does Java raise this compilation error ? If we look at the IdentifierExpression.java class file, we will find this block :

...
if (field.isLocal()) {
            LocalMember local = (LocalMember)field;
            if (local.scopeNumber < ctx.frameNumber && !local.isFinal()) {
                env.error(where, "invalid.uplevel", id);
            }
            if (!vset.testVar(local.number)) {
                env.error(where, "var.not.initialized", id);
                vset.addVar(local.number);
            }
            local.readcount++;
        }
...

As stated (if (!vset.testVar(local.number)) {), the JDK checks (with testVar) if the variable is assigned (Vset's source code where we can find testVar code). If not, it raises the error var.not.initialized from a properties file :

...
javac.err.var.not.initialized=\
    Variable {0} may not have been initialized.
...

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