Common Java Errors

List of common Java error messages:

cannot find symbol

"Cannot find symbol" errors generally occur when you try to reference an undeclared variable in your code. Consider the following example:

    public class Test {
        public static void main(String[] args) {
            int a = 3;
            int b = 4;
            int c = 20;
        
            average = (a + b + c)/5.0;
            System.out.println(average);
        }
    }

 
    1 error found:
    File: Test.java  [line: 7]
    Error: Test.java:7: cannot find symbol
    symbol  : variable average
    location: class Test

Here, the variable average has not been declared-- you need to tell the compiler what the type of average is; for example:

    double average = (a + b + c)/5.0;

Secondly, sometimes this error occurs because you are trying to reference a method in your code, but forget to include the parentheses that indicate a reference to a method, even when there are no parameters. For example:

    public class Test {
        public static void main(String[] args) {
            my_method;
        }
    
        public static void my_method() {
            System.out.println("Hello, world!");
        }
    }

    
    1 error found:
    File: Test.java  [line: 7]
    Error: Test.java:7: cannot find symbol
    symbol  : variable my_method
    location: class Test

Here, the compiler is looking for a variable called my_method in the main method; instead, you want to initiate a method call tomy_method:

    public class Test {
        public static void main(String[] args) {
            my_method();
        }
    
        public static void my_method() {
            System.out.println("Hello, world!");
        }
    }

Thirdly, this error could also occur if you forget to import a Java package that you need to use. For example, consider the following program that reads in an integer from the user:

    public class Test {     
        public static void main(String[] args) {
            Scanner console = new Scanner(System.in);
            int n = console.nextInt();
        }
    }

    
    2 errors found:
    File: Test.java  [line: 3]
    Error: cannot find symbol
      symbol:   class Scanner
      location: class Test
    File: Test.java  [line: 3]
    Error: cannot find symbol
      symbol:   class Scanner
      location: class Test

The issue here is that the program must import java.util.Scanner (or, more generally, java.util.*). Otherwise, the compiler does not know what a Scanner type is. You may encounter a similar error if you forget to import java.util.Arrays orjava.io.* when working with file input/output. To fix the code above:

    import java.util.*; // or --> import java.util.Scanner;    
    public class Test {     
        public static void main(String[] args) {
            Scanner console = new Scanner(System.in);
            int n = console.nextInt();
        }
    }

Finally, this error can occur if you have case-sensitive errors with variables. All identifiers in Java are case sensitive. This means that if you declare a variable named average and try to later refer to it using Average, the compiler will complain that it cannot find a symbol named Average.

expected

Errors of the form " expected" happen when the compiler detects a missing character in your code. The error message will tell you which character is missing and on which line. Consider the following program:

    public class Test
        public static void main(String[] args) {
            my_method();
        }
    
        public static void my_method() {
            System.out.println("Hello, world!")     
        }
    }
 
    
    2 errors found:
    File: Test.java  [line: 1]
    Error: Test.java:1: '{' expected
    File:.java  [line: 7]
    Error: Test.java:7: ';' expected

These error messages tell you there is a missing curly brace on the first line and a missing semi-colon on the seventh line. To fix this kind of error, simply place the missing character in the correct position in the code:

    public class Test {
        public static void main(String[] args) {
            my_method();
        }
    
        public static void my_method() {
            System.out.println("Hello, world!");   
        }
    }

expected

This error occurs when code is written outside of a method; it is typically caused by a mistake in curly braces. Consider the following example:

    public class Test { 
        System.out.println("Hello!");
    
        public static void main(String[] args) {
            System.out.println("World!");
        }
    }

 
    2 errors found:
    File: Test.java  [line: 2]
    Error:  expected
    File: Test.java  [line: 2]
    Error: illegal start of type

In this case, it is somewhat clear that the first print statement must be inside the main method for the code to compile. However, when there is more than one method and a curly brace error, the " expected" error can be harder to see:

    public class Test {     
        public static void main(String[] args) {
            System.out.println("Hello!");}
            System.out.println("World!");
        }
    }

 
    3 errors found:
    File: Test.java  [line: 4]
    Error:  expected
    File: Test.java  [line: 4]
    Error: illegal start of type
    File: Test.java  [line: 6]
    Error: class, interface, or enum expected

There is an extra curly brace in the code above, but the code is not properly indented so it is difficult to see. The effect of this is to end the main method immediately after the line that prints "Hello!," which leaves the print statement that prints "World!" outside of any method. To fix the error above, simply remove the curly brace at the end of the third line:

    public class Test {     
        public static void main(String[] args) {
            System.out.println("Hello!");
            System.out.println("World!");
        }
    }

invalid method declaration; return type required

Every method in Java requires that you explicitly state the return type of the method. Even methods that do not return a value must explicitly say void in the method signature, just as the main method does.When a method declaration does not contain a return type, this error will occur:

    public class Test {
        public static void main(String[] args) {
            int x = getValue();
            System.out.println(x);
        }
    
        public static getValue() {
            return 10;
        }
    }

 
    1 error found:
    File: Test.java  [line: 7]
    Error: Test.java:7: invalid method declaration; return type required

To fix this, simply insert the appropriate return type in the method signature:

    public class Test {
        public static void main(String[] args) {
            int x = getValue();
            System.out.println(x);
        }
    
        public static int getValue() {
            return 10;
        }
    }

java.lang.ArrayIndexOutOfBoundsException:

An ArrayIndexOutOfBoundsException is thrown when an attempt is made to access an index in an array that is not valid. The only valid indices for an array arr are in the range [0, arr.length - 1]; any attempt to access an index outside of this range will result in this error. For example:
```
    public class Test {
        public static void main(String[] args) {
            int[] arr = {1, 2, 3};
            for (int i = 0; i <= arr.length; i++) {
                System.out.println(arr[i]);
            }
        }
    }
 
    
    java.lang.ArrayIndexOutOfBoundsException: 3
 at Test.main(Test.java:5)
 at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
 at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:57)
 at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:43)
 at java.lang.reflect.Method.invoke(Method.java:606)
 at edu.rice.cs.drjava.model.compiler.JavacCompiler.runCommand(JavacCompiler.java:272)
 
```
This error is quite similar to the
java.lang.StringIndexOutOfBoundsException: String index out of range: error that we have previously discussed. The error message for this kind of error is similarly irrelevant toward the end of the message. However, the first line lets you know that a problem with an array index was encountered, and the index in error was 3, in this case. The next line tells you that it encountered this error on line 5 of Test.java, inside the main method.In this case, the error occurred because the for loop iterates too many times; the value of the loop index, i, reaches 4 and is therefore out of bounds. Instead, the upper bound should use the < boolean operator, or an equivalent statement.
```
    public class Test {
        public static void main(String[] args) {
            int[] arr = {1, 2, 3};
            for (int i = 0; i < arr.length; i++) {
                System.out.println(arr[i]);
            }
        }
    }
```
When dealing with an ArrayIndexOutOfBoundsException, it is usually helpful to print out the value of the index variable that is accessing the array and try to trace through to code to find out why it is reaching that (invalid) value.

java.lang.StringIndexOutOfBoundsException: String index out of range: v

A StringIndexOutOfBoundsException is thrown when an attempt is made to access an index in the String that is not valid. The only valid indices for a String str are in the range [0, str.length() - 1]; any attempt to access an index outside of this range will result in this error. For example:
```
    public class Test {
        public static void main(String[] args) {
            String str = "Hello, world!";

            String a = str.substring(-1, 3);
            String b = str.charAt(str.length());
            String c = str.substring(0, 20);
        }
    }
 
    
    java.lang.StringIndexOutOfBoundsException: String index out of range: -1
        at java.lang.String.substring(Unknown Source)
        at Test.main(Test.java:5)
        at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method)
        at sun.reflect.NativeMethodAccessorImpl.invoke(Unknown Source)
        at sun.reflect.DelegatingMethodAccessorImpl.invoke(Unknown Source)
        at java.lang.reflect.Method.invoke(Unknown Source)
        at edu.rice.cs.drjava.model.compiler.JavacCompiler.runCommand(JavacCompiler.java:271)
 
```
The error message for this kind of error becomes less relevant towards the end. However, the first line lets you know that a problem with a String index was encountered, and the index in error was -1. The next line tells you that it encountered this error while trying to perform the substring routine, which was called from the Test class on line 5. This "backtrace" of the error tells you the line numbers of the method calls involved so that you can trace your error to the source and correct it.Note that all of a, b, and c would have thrown this error, but the program was halted after the first occurred.
This is not a compile-time error, but rather a runtime error. In other words, the compiler will accept this kind of error because it is a logical error. Additionally, it may not be known before the program is run that the error will occur. To fix this error, you often have to correct the logic of your program to ensure that the program will not try to access an invalid index.

method in class cannot be applied to given types

This method occurs when you try to call a method using the wrong number or wrong order of parameters. For example, consider the following program:

    public class Test {     
        public static void main(String[] args) {
            myMethod(1.0, 2, "Hello!");
        }
    
        public static void myMethod(double d, String s, int x) {
            System.out.println(s + " " + d + " " + x);
        }
    }
    
 
    1 error found:
    File: Test.java  [line: 3]
    Error: method myMethod in class Test cannot be applied to given types;
      required: double,java.lang.String,int
      found: double,int,java.lang.String
      reason: actual argument int cannot be converted to java.lang.String by method invocation conversion

The error message for this error is very helpful. The line that says "required" tells you about what the method is expecting. It lists the order of the arguments that are required. In the example above, the parameters for myMethod should be a double, then aString, and then an int.The next line of the error message (which says "found") tells you what you (incorrectly) tried to use to call the method. In this example, we invoked the method using a double, then an int, and then a String-- which is the wrong order!
We can fix this by ensuring that the number and ordering of the method parameters is correct:

    public class Test {     
        public static void main(String[] args) {
            myMethod(1.0, "Hello!", 2);
        }
    
        public static void myMethod(double d, String s, int x) {
            System.out.println(s + " " + d + " " + x);
        }
    }

missing return statement

This method happens when you declare that a method will return a value, but then fail to actually return the value. For example:

    public class Test {
        public static void main(String[] args) {
            int x = twice(5);
            System.out.println(x);
        }
    
        public static int twice(int x) {
            int value = 2 * x;
        }
    }
 
     
    1 error found:
    File: Test.java  [line: 9]
    Error: Test.java:9: missing return statement

We have informed the compiler that the twice method will return an int, but we are missing the return statement:

    public class Test {
        public static void main(String[] args) {
            int x = twice(5);
            System.out.println(x);
        }
    
        public static int twice(int x) {
            int value = 2 * x;
            return value;
        }
    }

Returning values in if statements can also trick the compiler into thinking that it's possible no value will be returned, as in the following example:

    public class Test {
        public static void main(String[] args) {
            int x = absVal(-5);
            System.out.println(x);
        }
    
        public static int absVal(int x) {
            if (x < 0) {
                return -x;
            }
   
            if (x >= 0) {
                return x;
            }
        }
    }
 
     
    1 error found:
    File: Test.java  [line: 15]
    Error: Test.java:15: missing return statement

To avoid this, we can either use an else statement (as we did in the Variable might not have been initialized example), or we can simply not use the second if statement because we know that if we've reached that point in the method we can just return x:

    public class Test {
        public static void main(String[] args) {
            int x = absVal(-5);
            System.out.println(x);
        }
    
        public static int absVal(int x) {
            if (x < 0) {
                return -x;
            }
   
            return x;
        }
    }

unreachable statement

An "unreachable statement" error occurs when the compiler detects that it is impossible to reach a given statement during the flow of a program. This error is often caused by placing statements after return or break. For example:

    public class Test {
        public static void main(String[] args) {
            int value = twice(5);
            System.out.println(value);
        }
    
        public static int twice(int x) {
            int twice = 2 * x;
            return twice;
            System.out.println("Returning " + twice);
        }
    }
 
    2 errors found:
    File: Test.java  [line: 10]
    Error: Test.java:10: unreachable statement
    File: Test.java  [line: 11]
    Error: Test.java:11: missing return statement

The compiler gives two errors: one to indicate that the line System.out.println("Returning " + twice); is an unreachable statement, and another because it assumes that if we can get to that print statement, then we would need a returnstatement somewhere after it.We can fix this by placing the print statement before the return so it can be executed:

    public class Test {
        public static void main(String[] args) {
            int value = twice(5);
            System.out.println(value);
        }
    
        public static int twice(int x) {
            int twice = 2 * x;
            System.out.println("Returning " + twice);
            return twice;
        }
    }

variable might not have been initialized

This error occurs when the compiler believes you're trying to use a variable that has not been "initialized"-- or given an initial value-- yet. In a very simple case:

    public class Test {
        public static void main(String[] args) {
            int x = 2;
            int y;
            System.out.println(x + y);
        }
    } 

    
    1 error found:
    File: Test.java  [line: 5]
    Error: Test.java:5: variable y might not have been initialized

Here, you have not told the compiler what the value of y is. Therefore, y cannot be printed; it needs to be initialized as x is in this example.In more complicated scenarios, if statements can cause this error if you are not careful about ensuring that a variable is initialized. For example:

    public class Test {
        public static void main(String[] args) {
            int x;
            boolean setX = false;
   
            if (setX) {
                x = 10;
            }
   
            System.out.println(x);
        }
    }

    
    1 error found:
    File: Test.java  [line: 8] 
    Error: Test.java:8: variable x might not have been initialized

Here again it is clear that the variable x will not be initialized, and therefore an error occurs. However, cases can also arise where it is clear to us that one of the cases has to be reached and therefore the error should not happen. However, the compiler is not always smart enough to see cases that we as humans can see. For example:

    public class Test {
        public static void main(String[] args) {
            int x;
            boolean setToTen = false;
         
            if (setToTen) {
                x = 10;
            }
        
            if (!setToTen) {
                x = 0;
            }
        
            System.out.println(x);
        }
    }

    
    1 error found:
    File: Test.java  [line: 14]
    Error: Test.java:14: variable x might not have been initialized

It may appear obvious that x will get set to a value one way or another, but the compiler cannot see this. One way to fix this error is to use an else statement. When using an else statement, the compiler is smart enough to see that in at least one case x will be initialized:

    public class Test {
        public static void main(String[] args) {
            int x;
            boolean setToTen = false;
        
            if (setToTen) {
                x = 10;
            } else {
                x = 0;
            }
        
            System.out.println(x);
        }
    }

Original document created by Cody Doucette '14.

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