Variable might not have been initialized error

When I try to compile this:

public static Rand searchCount (int[] x)
{
    int a ;
    int b ;

    ...

    for (int l= 0; l<x.length; l++)
    {
        if (x[l] == 0)
        a++ ;
        else if (x[l] == 1)
        b++ ;
    }

    ...

}

I get these errors:

Rand.java:72: variable a might not have been initialized
                a++ ;
                ^
Rand.java:74: variable b might not have been initialized
                b++ ;
                ^
2 errors

It seems to me that I initialized them at the top of the method. What’s going wrong?

12 Answers

You declared them, but you didn’t initialize them. Initializing them is setting them equal to a value:

int a;        // This is a declaration
a = 0;        // This is an initialization
int b = 1;    // This is a declaration and initialization

You get the error because you haven’t initialized the variables, but you increment them (e.g., a++) in the for loop.

Java primitives have default values but as one user commented below

Their default value is zero when declared as class members. Local variables don’t have default values

Local variables do not get default values. Their initial values are undefined with out assigning values by some means. Before you can use local variables they must be initialized.

There is a big difference when you declare a variable at class level (as a member ie. as a field) and at method level.

If you declare a field at class level they get default values according to their type. If you declare a variable at method level or as a block (means anycode inside {}) do not get any values and remain undefined until somehow they get some starting values ie some values assigned to them.

If they were declared as fields of the class then they would be really initialized with 0.

You’re a bit confused because if you write:

class Clazz {
  int a;
  int b;

  Clazz () {
     super ();
     b = 0;
  }

  public void printA () {
     sout (a + b);
  }

  public static void main (String[] args) {
     new Clazz ().printA ();
  }
}

Then this code will print “0”. It’s because a special constructor will be called when you create new instance of Clazz. At first super () will be called, then field a will be initialized implicitly, and then line b = 0 will be executed.

You declared them, but not initialized.

int a; // declaration, unknown value
a = 0; // initialization
int a = 0; // declaration with initialization

You declared them, but you didn’t initialize them with a value. Add something like this:

int a = 0;

You declared them but did not provide them with an intial value – thus, they’re unintialized. Try something like:

public static Rand searchCount (int[] x)  
{ 
  int a = 0 ;  
  int b = 0 ; 

and the warnings should go away.

You declared them at the start of the method, but you never initialized them. Initializing would be setting them equal to a value, such as:

int a = 0;
int b = 0;

Imagine what happens if x[l] is neither 0 nor 1 in the loop. In that case a and b will never be assigned to and have an undefined value. You must initialize them both with some value, for example 0.

It’s a good practice to initialize the local variables inside the method block before using it. Here is a mistake that a beginner may commit.

  public static void main(String[] args){
    int a;
    int[] arr = {1,2,3,4,5};
    for(int i=0; i<arr.length; i++){
        a = arr[i];
    }
    System.out.println(a);
  }

You may expect the console will show ‘5’ but instead the compiler will throw ‘variable a might not be initialized’ error. Though one may think variable a is ‘initialized’ inside the for loop, the compiler does not think in that way. What if arr.length is 0? The for loop will not be run at all. Hence, the compiler will give variable a might not have been initialized to point out the potential danger and require you to initialize the variable.

To prevent this kind of error, just initialize the variable when you declare it.

int a = 0;

You haven’t initialised a and b, only declared them. There is a subtle difference.

int a = 0;
int b = 0;

At least this is for C++, I presume Java is the same concept.

Set variable “a” to some value like this,

a=0;

Declaring and initialzing are both different.

Good Luck

Since no other answer has cited the Java language standard, I have decided to write an answer of my own:

In Java, local variables are not, by default, initialized with a certain value (unlike, for example, the 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)).

Therefore, since the variables a and b are not initialized :

 for (int l= 0; l<x.length; l++) 
    {
        if (x[l] == 0) 
        a++ ;
        else if (x[l] == 1) 
        b++ ;
    }

the operations a++; and b++; could not produce any meaningful results, anyway. So it is logical for the compiler to notify you about it:

Rand.java:72: variable a might not have been initialized
                a++ ;
                ^
Rand.java:74: variable b might not have been initialized
                b++ ;
                ^

However, one needs to understand that the fact that a++; and b++; could not produce any meaningful results has nothing to do with the reason why the compiler displays an error. But rather because it is explicitly set on the Java language specification that

A local variable (§14.4, §14.14) must be explicitly given a value (…)

To showcase the aforementioned point, let us change a bit your code to:

public static Rand searchCount (int[] x) 
{
    if(x == null || x.length  == 0)
      return null;
    int a ; 
    int b ; 

    ...   

    for (int l= 0; l<x.length; l++) 
    {
        if(l == 0)
           a = l;
        if(l == 1)
           b = l;
    }

    ...   
}

So even though the code above can be formally proven to be valid (i.e., the variables a and b will be always assigned with the value 0 and 1, respectively) 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 variables a and b are initialized according to the local variable initialization rules, and reacts accordingly (e.g., displaying a compilation error).

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