CODEWITHSUNDEEP
javavariablescompiler-errorslocalinitialization
Ivan
Ivan

Reputation: 327

Java: "Local variable may not have been initialized" not intelligent enough?

Consider the following method:

void a ()
{
    int x;
    boolean b = false;
    if (Math.random() < 0.5)
    {
        x = 0;
        b = true;
    }
    if (b)
        x++;
}

On x++ I get the "Local variable may not have been initialized" error. Clearly x will never be used uninitialized. Is there any way to suppress the warning except by initializing x? Thanks.

Upvotes: 4

Views: 26394

Answers (4)

Codebling
Codebling

Reputation: 11397

You can and should be defining the value of x unconditionally if it will be used later in your code.

There are a few ways to do this:

On initialization

int x = 0;

Because this is outside the conditional (if), Java won't complain.

Add else clause to conditional

if (Math.random() < 0.5)
{
    x = 0;
    b = true;
} else 
{
    x = 1;
}

Because there is an else to this if, and both code paths initialize x, Java will also be happy with this.

Move your usage of the variable into the conditional block

Clearly the question has a minimally-reproducible example, not a full one, but if you only ever want to use the variable conditionally, then it belongs in the conditional block.

if (Math.random() < 0.5)
{
    x = 0;
    x++;
}

If you don't aren't conditionally using the variable, then you need to provide an integer value to use in case Math.random() >= 0.5, using one of the solutions above.

Upvotes: 0

Keppil
Keppil

Reputation: 46239

No, there is no way Java can examine all possible code paths for a program to determine if a variable has been initialized or not, so it takes the safe route and warns you.

So no, you will have to initialize your variable to get rid of this.

Upvotes: 6

Denys S&#233;guret
Denys S&#233;guret

Reputation: 382464

There is one :

void a () {
    if (Math.random() < 0.5) {
        int x = 1;
    }
}

The compiler isn't responsible for devising and testing the algorithm. You are.

But maybe you should propose a more practical use case. Your example doesn't really show what's your goal.

Upvotes: 2

Ankur
Ankur

Reputation: 12774

Why don't you simply use

void a ()
{
    int x;
    boolean b = false;
    if (Math.random() < 0.5)
    {
        x = 0;
        b = true;
        x++;
    }
    if (b) {
        //do something else which does not use x
    }
}

In the code why do you want to use x outside the first if block, all the logic involving x can be implemented in the first if block only, i don't see a case where you would need to use the other if block to use x.

EDIT: or You can also use:

void a ()
{
    int x;
    boolean b = (Math.random() < 0.5);
    if (b) {
         x=1
        //do something 
    }
}

Upvotes: 1

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