How to fix "variable might not have been initialized" error in Java? Example

This error occurs when you are trying to use a local variable without initializing it. You won't get this error if you use an uninitialized class or instance variable because they are initialized with their default value like Reference types are initialized with null and integer types are initialized with zero, but if you try to use an uninitialized local variable in Java, you will get this error. This is because Java has the rule to initialize the local variable before accessing or using them and this is checked at compile time. If the compiler believes that a local variable might not have been initialized before the next statement which is using it, you get this error. You will not get this error if you just declare the local variable but will not use it.

Let's see a couple of examples:

public class Main {

  public static void main(String[] args) {

    int a = 2;
    int b;
    int c = a + b;

  }

}

You can see we are trying to access variable "b" which is not initialized in statement c = a + b, hence when you run this program in Eclipse, you will get the following error:

Exception in thread "main" java.lang.Error: Unresolved compilation problem:
The local variable b may not have been initialized
at Main.main(Main.java:14)

The error message is very clear, it's saying that local variable "b" has not initialized until line 14, where it has been read, which violates the Java rule of initializing the local variable before use.

Though this rule is only for local variables, if you try to access uninitialized member variables e.g. a static or non-static variable, you will not get this error as shown below:

public class Main {

  private static int total;
  private int sum;

  public static void main(String[] args) {

    int d = total; // no error because total is static variable
    Main m = new Main();
    int e = m.sum; // no error bcasue sum is instnace variable
  }

}

This program will both compile and run fine.




How to fix "variable might not have been initialized" error in Java? Example

Now, there are some tricky scenarios where you think that you have initialized the variable but the compiler thinks otherwise and throws a "variable might not have been initialized" error. One of them is creating more than one local variable in the same line as shown in the following Java program:

public class Main {

  public static void main(String[] args) {

    int a, b = 0;
    System.out.println("a:" + a);
    System.out.println("b:" + b);

  }

}

Here you might think that both variables "a" and "b" are initialized to zero but that's not correct. Only variable b is initialized and variable "a" is not initialized, hence when you run this program, you will get the "variable might not have been initialized" error as shown below:

Exception in thread "main" java.lang.Error: Unresolved compilation problem: 
The local variable a may not have been initialized

at Main.main(Main.java:13)

Again, the error message is very precise, it says that variable "a" is not initialized on line 13 where you have used it for reading its value. See these best Java Programming tutorials to learn more about how variables are initialized in Java.

One more scenario, where the compiler complains about "variable might not have been initialized" is when you initialize the variable inside if() block since it is a condition block, compiler know that variable may not get initialized when if block is not executed, so it complains as shown in the following program:

public class Main {

  public static void main(String[] args) {
    int count;

    if (args.length > 0) {
      count = args.length;
    }

    System.out.println(count);

  }

}

In this case, the variable count will not be initialized before you use it on System.out.println() statement if args.length is zero, hence compiler will throw "variable might not have been initialized" when you run this program as shown below:

Exception in thread "main" java.lang.Error: Unresolved compilation problem: 
The local variable count may not have been initialized

at Main.main(Main.java:17)


Now, sometimes, you will think that compiler is wrong because you know that variable is always going to be initialized but in reality, compilers are not as smart as you and you see this error as shown in the following program:

public class Main{

  public static void main(String[] args) {
    int count;

    if (args.length > 0) {
      count = args.length;
    }

    if (args.length == 0) {
      count = 0;
    }

    System.out.println(count);

  }

}

How to fix "variable might not have been initialized" error in Java

Now, you know that count will always initialize because the length of the argument array would either be zero or greater than zero, but the compiler is not convinced and it will throw the "variable might not have been initialized" error as shown below:

Exception in thread "main" java.lang.Error: Unresolved compilation problem: 
The local variable count may not have been initialized

at Main.main(Main.java:21)


One way to convince the compiler is to use the else block, this will satisfy the compiler and the error will go away as shown below:

public class Main{

  public static void main(String[] args) {
    int count;

    if (args.length > 0) {
      count = args.length;
    } else {
      count = 0;
    }

    System.out.println(count);

  }

}

If you run this program, there won't be any compile error because now the compiler knows for sure that count will be initialized before accessed. If you remove all the if-else and System.out.println() block then also code will compile fine because we have only declared the count variable and never used it. You can also see these free Java tutorials to learn more about rules related to initializing local, instance, and class variables.

There are more scenarios where you get the "variable might not have been initialized" error, especially when you initialize a variable inside a block e.g. try or catch block. So beware of this rule, it's not a big problem but sometimes becomes a headache for Java beginners, especially when they get tons of "variable might not have been initialized" errors when they compile their Java source file.


14 comments:

  1. man u just did it love u man . thnks .for this

    ReplyDelete
  2. Why does this code complain about rc not being initialized on the return?

    public static String testInitializeInLoop()
    throws Exception
    {
    String rc;

    for (int i = 0; i < 2; i++) {

    try {

    if (i == 0) {
    throw new Exception("failed to connect");
    }
    rc = "done";

    } catch (Exception e) {
    if (i+1 < 2) {
    System.out.println("Failed on one - " + e);
    continue;

    } else {
    String msg = "failed on both - " + e;
    System.out.println(msg);
    throw new Exception(msg);
    }
    }
    }

    return rc;
    }

    ReplyDelete
    Replies
    1. because you are initializing rc inside loop, just do String rc="" and it will not give any error. A local variable must initialize in all path. Complier things that if control will not go inside for then rc will never initialize.

      Delete







  3. import java.util.Scanner;

    public class MyFifthClass {
    public static void main(String[] args) {
    Scanner keyboard = new Scanner(System.in);
    String choice;
    int x, y, result;


    choice = keyboard.nextLine();
    x = keyboard.nextInt();
    y = keyboard.nextInt();

    switch (choice) {
    case “1”:
    result = x + y;
    break;

    case “2”:
    result = y - x;
    break;

    case “3”:
    result = x * y;
    break;
    }

    System.out.println(“Answer: ” + result);
    }
    }




    major help please result isnt initialized it says

    ReplyDelete
  4. here the error is with variable 'A' in case 4 , how can I rectify it??/

    import java . util.*;
    class all_in_ONE
    {void tri_angle_ie( )
    {Scanner x=new Scanner(System.in);
    System.out.println("enter three side of triangle ");
    int a=x.nextInt();
    int b=x.nextInt();
    int c=x.nextInt();
    System.out.println("greeting, if you like calculate ___ enter ____:-");
    System.out.println("area - 1\nperimeter - 2\nheight - 3\nvolume - 4");
    int d=x.nextInt();
    double s=(a+b+c)/2.0;
    switch (d)
    { case 1://area of triangle
    double e=s*(s-a)*(s-b)*(s-c);
    int f= (int)Math.round(e);
    int A= (int)Math.sqrt(f);
    System.out.println("the area of the triangle is= "+A);
    break;
    case 2://perimeter
    int S= a+b+c;
    System.out.println("the perimeter of the triangle is= "+S);
    break;
    case 3: //height
    double H;
    if(a<b)
    {H=Double.valueOf(b);}
    else
    { if(c<a)
    {H=Double.valueOf(a);}
    else
    { H =Double.valueOf(c);}}
    double h = 2*(A/H);
    System.out.println("the height of the trianle is= "+h);
    break;
    case 4: //volume
    int V= (int)Math.round(h);
    int v=A*V;
    System.out.println("the volume of the trianle is= "+v);
    break;
    }
    }
    }

    ReplyDelete
    Replies
    1. A is a local variable which means its only visible in the case 1 score where it is declared. If you want to make A visible in case 4, declare it outside case 1, above the switch statement.

      Delete
  5. class Sample2 {

    public static void main(String args[]) {

    //Intialize the char variable below with operators (+, -, *, /) one by one and check the output
    char operator = ( '+' );

    double first =10, second =5;
    double result;

    switch (operator) {
    case '+':
    // code to perform addition of first and second
    result = first + second;
    break;

    case '-':
    // code to perform subtraction

    result = first - second;

    break;

    case '*':
    // code to perform multiplication
    result = first*second ;
    break;
    case '/':
    // code to perform division
    result = first/ second;

    //If operator doesn't match any case constant (+, -, *, /)
    default:
    // Print Error stating operator is not correct

    }

    System.out.println( first+ " " + operator + " " + second + "=" + result);
    }
    }



    it showing variable result might not have been initialized how can i sir can u help me

    ReplyDelete
  6. instead of doing double result; just do double result = 0; and it should be good. The error is coming because you are initializing it on case statement which means if code doesn't goes there the result variable will not be initialized.

    ReplyDelete
  7. thank you, this helped a lot!

    ReplyDelete
  8. package Assignment.Q056;
    import java.util.Random;
    import java.util.Scanner;

    public class Q056Test
    {
    public static void main(String[] args)
    {
    Scanner input = new Scanner(System.in);
    Q056 test = new Q056();

    int correctAnswer, answer, correct = 0, correct2;

    String choice;

    do
    {
    correctAnswer = test.question();

    do
    {
    System.out.print("\nYour answer: ");
    answer = input.nextInt();

    if (answer == correctAnswer)
    {
    int randomInt = (int)(10.0 * Math.random());
    randomInt = randomInt%4 + 1;

    switch(randomInt)
    {
    case 1 :
    System.out.println("Very good!");
    break;
    case 2 :
    System.out.println("Nice Work!");
    break;
    case 3 :
    System.out.println("Excellent!");
    break;
    case 4 :
    System.out.println("Keep up the good work!");
    break;
    default :
    System.out.println("Default!");
    break;
    }
    }
    else
    {
    int randomInt = (int)(10.0 * Math.random());
    randomInt = randomInt%4 + 1;

    switch(randomInt)
    {
    case 1 :
    System.out.println("No. Please try again.");
    break;
    case 2 :
    System.out.println("Wrong. Try once more.");
    break;
    case 3 :
    System.out.println("Don’t give up!");
    break;
    case 4 :
    System.out.println("No. Keep trying.");
    break;
    default :
    System.out.println("Default!");
    break;
    }
    }
    } while (answer != correctAnswer);

    System.out.println("Do you want continue? ('1 = yes' or '2 = no') ");
    choice = input.next();
    }
    while (choice.equalsIgnoreCase("1"));

    for(int i = 0; i < 1; i++)
    {
    if(choice.equals(correctAnswer))
    correct++;
    {
    System.out.println("Correct answers : "+ correct);
    }

    }

    correct2 = correct*10;
    if(correct2 <= 75)

    System.out.println("Please ask your teacher for extra help");

    else if(correct2 >= 75)

    System.out.println("Congratulations! you are ready to go to the next level");

    }
    }

    Sir may I know why when I run the code at the end it shows :

    correct answers: 0
    Please ask your teacher for extra help

    instead of calculating the real number of correct answer:

    ReplyDelete
  9. Thank you, I started learning Java recently and this helped me a lot!

    ReplyDelete
  10. hello sir/mam;
    Actually I'm the beginner in the java programming and i got same error last 1 hr. I try to resolve it but not get to resolved if I initialized to 0 then the area we get is 0.

    import java.util.Scanner;
    class arrow{
    void setDim() {
    Scanner sc = new Scanner(System.in);
    System.out.println("Enter length: ");
    int length = sc.nextInt();
    System.out.println("Enter breadth: ");
    int breadth = sc.nextInt();
    }
    void getArea() {
    int length;
    int breadth;

    int arrow = length * breadth;
    System.out.println("Area of Rectangle is: "+arrow);
    }
    }
    public class Main{

    public static void main(String[] args) {
    arrow a1 = new arrow();
    a1.setDim();
    a1.getArea();
    }
    }


    please help me to resolve this error and also provide some study material......

    ReplyDelete

Feel free to comment, ask questions if you have any doubt.