Error not a statement java

Being a java developer, you must encounter numberless bugs and errors on daily basis. Whether you are a beginner or experienced software engineers, errors are inevitable but over time you can get experienced enough to be able to correct them efficiently. One such very commonly occurring error is “Illegal start of expression Java error”.

The illegal start of expression java error is a dynamic error which means you would encounter it at compile time with “javac” statement (Java compiler). This error is thrown when the compiler detects any statement that does not abide by the rules or syntax of the Java language. There are numerous scenarios where you can get an illegal start of expression error. Missing a semicolon at the end of The line or an omitting an opening or closing brackets are some of the most common reasons but it can be easily fixed with slight corrections and can save you a lot of time in debugging.

Following are some most common scenarios where you would face an illegal start of expression Java error along with the method to fix them,

1. Use of Access Modifiers with local variables

Variables that are declared inside a method are called local variables. Their functionality is exactly like any other variable but they have very limited scope just within the specific block that is why they cannot be accessed from anywhere else in the code except the method in which they were declared.

Access modifier (public, private, or protected) can be used with a simple variable but it is not allowed to be used with local variables inside the method as its accessibility is defined by its method scope.

See the code snippet below,

1.	public class classA {
2.	    public static void main(String args[])
3.	    {        
4.	       public int localVar = 5;
5.	    }
6.	}

 Here the public access modifier is used with a local variable (localVar).

This is the output you will see on the terminal screen:

$javac classA.java
 
classA.java:4: error: illegal start of expression
       public int localVar = 5;
       ^
1 error

It reports 1 error that simply points at the wrong placement of access modifier. The solution is to either move the declaration of the local variable outside the method (it will not be a local variable after that) or simply donot use an access modifier with local variables.

2. Method Inside of Another Method

Unlike some other programming languages, Java does not allows defining a method inside another method. Attempting to do that would throw the Illegal start of expression error.

Below is the demonstration of the code:

1.	public class classA {
2.	    public static void main(String args[]) {        
3.	       int localVar = 5;
4.	       public void anotherMethod(){ 
5.	          System.out.println("it is a method inside a method.");
6.	       }
7.	    }
8.	}

This would be the output of the code,

 $javac classA.java
 
classA.java:5: error: illegal start of expression
       public void anotherMethod()
       ^
classA.java:5: error: illegal start of expression
       public void anotherMethod()
              ^
classA.java:5: error: ';' expected
       public void anotherMethod()
                                ^
3 errors

It is a restriction in Java so you just have to avoid using a method inside a method to write a successfully running code. The best practice would be to declare another method outside the main method and call it in the main as per your requirements.

3. Class Inside a Method Must Not Have Modifier

Java allows its developers to write a class within a method, this is legal and hence would not raise any error at compilation time. That class will be a local type, similar to local variables and the scope of that inner class will also be restricted just within the method. However, an inner class must not begin with access modifiers, as modifiers are not to be used inside the method.

In the code snippet below, the class “mammals” is defined inside the main method which is inside the class called animals. Using the public access modifier with the “mammals” class will generate an illegal start of expression java error.

1.	public class Animals {
2.	    public static final void main(String args[]){        
3.	      public class mammals { }
4.	    }
5.	}

 The output will be as follows,

$javac Animals.java
 
Animals.java:4: error: illegal start of expression
       public class mammals { }
       ^
1 error

This error can be fixed just by not using the access modifier with the inner class or you can define a class inside a class but outside of a method and instantiating that inner class inside the method.

Below is the corrected version of code as well,

1.	class Animals {
2.	   
3.	   // inside class
4.	   private class Mammals {
5.	      public void print() {
6.	         System.out.println("This is an inner class");
7.	      }
8.	   }
9.	   
10.	   // Accessing the inside class from the method within
11.	   void display_Inner() {
12.	      Mammals inside = new Mammals();
13.	      inside.print();
14.	   }
15.	}
16.	public class My_class {
17.	 
18.	   public static void main(String args[]) {
19.	      // Instantiating the outer class 
20.	      Animals classA = new Animals();
21.	      // Accessing the display_Inner() method.
22.	      classA.display_Inner();
23.	   }
24.	}

 Now you will get the correct output,

$javac Animals.java
 
$java -Xmx128M -Xms16M Animals
 
This is an inner class

4.Nested Methods

Some recent programming languages, like Python, supports nested methods but Java does not allow to make a method inside another method.  You will encounter the illegal start of expression java error if you try to create nested methods.

Below mentioned is a small code that attempts to declare a method called calSum inside the method called outputSum,

1.	public class classA {
2.	    public void outputSum(int num1, int num2) {
3.	        System.out.println("Calculate Result:" + calSum(x, y));
4.	        public int calSum ( int num1, int num2) {
5.	            return num1 + num2;
6.	        }
7.	    }
8.	}

 And here is the output,

$ javac classA.java
NestedMethod.java:6: error: illegal start of expression
        public int calSum ( int num1, int num2) {
        ^
classA.java:6: error: ';' expected
        public int calSum ( int num1, int num2) {
                          ^
classA.java:6: error:  expected
        public int calSum ( int num1, int num2) {
                                   ^
NestedMethod.java:6: error: not a statement
        public int calSum ( int num1, int num2) {
                                           ^
NestedMethod.java:6: error: ';' expected
        public calSum ( int num1, int num2) {
                                         ^
5 errors

The Java compiler has reported five compilation errors. Other 4 unexpected errors are due to the root cause. In this code, the first “illegal start of expression” error is the root cause. It is very much possible that a single error can cause multiple further errors during compile time. Same is the case here. We can easily solve all the errors by just avoiding the nesting of methods. The best practice, in this case, would be to move the calSum() method out of the outputSum() method and just call it in the method to get the results.

See the corrected code below,

1.	public class classA {
2.	    public void outputSum(int num1, int num2) {
3.	        System.out.println("Calculation Result:" + calSum(x, y));
4.	    }
5.	    public int calSum ( int num1, int num2) {
6.	        return x + y;
7.	    }
8.	}

5. Missing out the Curly “{ }“ Braces

Skipping a curly brace in any method can result in an illegal start of expression java error. According to the syntax of Java programming, every block or class definition must start and end with curly braces. If you skip any curly braces, the compiler will not be able to identify the starting or ending point of a block which will result in an error. Developers often make this mistake because there are multiple blocks and methods nested together which results in forgetting closing an opened curly bracket. IDEs usually prove to be helpful in this case by differentiating the brackets by assigning each pair a different colour and even identify if you have forgotten to close a bracket but sometimes it still gets missed and result in an illegal start of expression java error.

In the following code snippet, consider this class called Calculator, a method called calSum perform addition of two numbers and stores the result in the variable total which is then printed on the screen. The code is fine but it is just missing a closing curly bracket for calSum method which will result in multiple errors.

1.	public class Calculator{
2.	  public static void calSum(int x, int y) {
3.	    int total = 0;
4.	    total = x + y;
5.	    System.out.println("Sum = " + total);
6.	 
7.	  public static void main(String args[]){
8.	    int num1 = 3;
9.	    int num2 = 2;
10.	   calcSum(num1,num2);
11.	 }
12.	}

Following errors will be thrown on screen,

$javac Calculator.java
Calculator.java:12: error: illegal start of expression public int calcSum(int x, int y) { ^ 
Calculator.java:12: error: ';' expected 
 
Calculator.java:13: error: reached end of file while parsing
}
 ^
3 error

The root cause all these illegal starts of expression java error is just the missing closing bracket at calSum method.

While writing your code missing a single curly bracket can take up a lot of time in debugging especially if you are a beginner so always lookout for it.

6. String or Character Without Double Quotes “-”

Just like missing a curly bracket, initializing string variables without using double quotes is a common mistake made by many beginner Java developers. They tend to forget the double quotes and later get bombarded with multiple errors at the run time including the illegal start of expression errors.

If you forget to enclose strings in the proper quotes, the Java compiler will consider them as variable names. It may result in a “cannot find symbol” error if the “variable” is not declared but if you miss the double-quotations around a string that is not a valid Java variable name, it will be reported as the illegal start of expression Java error.

The compiler read the String variable as a sequence of characters. The characters can be alphabets, numbers or special characters every symbol key on your keyboard can be a part of a string. The double quotes are used to keep them intact and when you miss a double quote, the compiler can not identify where this series of characters is ending, it considers another quotation anywhere later in the code as closing quotes and all that code in between as a string causing the error.

Consider this code snippet below; the missing quotation marks around the values of the operator within if conditions will generate errors at the run time.

1.	public class Operator{
2.	  public static void main(String args[]){
3.	    int num1 = 10;
4.	    int num2 = 8;
5.	    int output = 0; 
6.	    Scanner scan = new Scanner(System.in);
7.	    System.out.println("Enter the operation to perform(+OR)");
8.	    String operator= scan.nextLine();
9.	    if(operator == +)
10.	  {
11.	     output = num1 + num2;
12.	  }
13.	  else if(operator == -)
14.	  {
15.	     output = num1 - num2;
16.	  }
17.	  else
18.	  {
19.	     System.out.prinln("Invalid Operator");
20.	  }
21.	  System.out.prinln("Result = " + output); 
22.	}

String values must be always enclosed in double quotation marks to avoid the error similar to what this code would return, 

$javac Operator.java
 
Operator.java:14: error: illegal start of expression
if(operator == +)
                ^
Operator.java:19: error: illegal start of expression
   if(operator == -)
                   ^
3 error

Conclusion

In a nutshell, to fix an illegal start of expression error, look for mistakes before the line mentioned in the error message for missing brackets, curly braces or semicolons. Recheck the syntax as well. Always look for the root cause of the error and always recompile every time you fix a bug to check as it could be the root cause of all errors.

See Also: Java Feature Spotlight – Sealed Classes

Such run-time errors are designed to assist developers, if you have the required knowledge of the syntax, the rules and restrictions in java programming and the good programming skills than you can easily minimize the frequency of this error in your code and in case if it occurs you would be able to quickly remove it.

Do you have a knack for fixing codes? Then we might have the perfect job role for you. Our careers portal features openings for senior full-stack Java developers and more.

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When a method is declared as having a return type of void, it cannot contain any return statements which return a value (it can, however, contain a return statement by itself, which will simply end the execution of the method). This problem is usually caused by accidentally making a method be of type void when it shouldn’t be or by accidentally including a return statement where there shouldn’t be one.

This error will most likely be caused when you omit the keyword class or interface, as seen in the example below.

Using TestInterface, consider the following example.

This error occurs when you have an else or else if clause without an if clause above it. This is most likely because you have accidentally forgotten/added extra { or } braces, or you have placed an else clause in the wrong place, as illustrated in the example below. The key idea is that every else clause needs to have an associated if clause which occurs before it.

Java is very specific about use of characters such as semicolons, brackets, or braces. Forgetting a semicolon is the simplest of these errors, and is fixed by placing a semicolon at the end of the line which causes the error. Brackets can be more complicated, because you may have to read through several nested if statements or loops in order to make sure that all brackets «match up» with each other properly. This is one of the reasons why indenting your code properly is a good idea.

An identifier is another term for a variable. In Java, every variable name must begin with a letter/underscore and can then have any combination of letters, numbers, and underscores. The example below illustrates two cases you may run into.

In a Java assignment statement, the type of the variable on the left hand side must be the same as the type of the variable on the right hand side (or it needs to be able to be cast first in order to make it work). The example below would give three ‘incompatible types’ error messages.

This error may also occur when trying to pass a value of one type into a method which expects another. An example of this is below.

You may also get a similar error if you are attempting to cast incorrectly. Recall that a primitive type may not be cast to an object or vice-versa. When casting between two objects, recall that object «A» may be cast to be of the same type as object «B» only if B’s class extends A’s class. As an example, consider a class Cat which extends the class Animal.

In Java, every method which does not have a return type of void must have at least one return statement.

This compile-time error can have a very subtle point which is often overlooked. If the return type of a method is not void, Java needs to ensure that the method will return a value in every possible case. That is, if all of your return statements are nested within if statements, Java will disallow the compilation process to continue because there is a chance that no return statement will be reached. The example below illustrates this.

In Java, recall that a method’s signature consists of the method’s name and the types of its actual parameters. For example, the method public void move(int howFar) has the signature move(int). When a method call is made, a method with a signature exactly matching that of the method call must exist (e.g. the method name, the number of parameters, and the types of all parameters must match exactly). This error is illustrated below.

Each line of Java code must have some sort of meaningful purpose. The compile-time error «not a statement» is usually the result of typing only part of a statement, leaving it incomplete and meaningless. One example of how this may occur (there are many) is seen below. In general, asking yourself «what was I trying to do with this line of code?» is sufficient to fix the problem.

For each method, Java requires a return type (e.g. String, int, boolean). Note that void is also considered to be a «return type» even though a method with a void return type does not return a value.

This error usually occurs if you have a statement placed directly after a return statement. Since Java executes code line-by-line (unless a method call is made), and since a return statement causes the execution of the program to break out of that method and return to the caller, then anything placed directly after a return statement will never be reached.

This compile-time error is typically caused by a programmer attempting to declare a variable twice. Recall that a statment such as int temp; declares the variable named temp to be of type int. Once this declaration has been made, Java can refer to this variable by its name (e.g. temp = 15;) but does not let the programmer declare it a second time by including the variable type before the name, as seen in the example below.

You may also get this error if you have tried to declare two variables of different types using the same variable name (in Java, every variable must have a unique name). The example below illustrates this using two variables of type int and String.

The only time where repetition of variable names or declarations are allowed is if the two variables are in different scopes. Recall that the scope of a variable is determined by the set of braces, { and }, in which it is enclosed. This is illustrated in the two examples below.

This error occurs when a variable is not declared within the same (or an «outer») scope in which it is used. Recall that the scope of a variable declaration is determined by its location with { and } braces. The example below illustrates a simple case of this error. To fix this problem, make sure that the variable is declared either in the same scope in which it is being used, or it being delcared in an «outer» scope.

Автор оригинала: Kai Yuan.

1. Обзор

“Незаконное начало выражения”-это распространенная ошибка, с которой мы можем столкнуться во время компиляции.

В этом уроке мы рассмотрим примеры, иллюстрирующие основные причины этой ошибки и способы ее устранения.

2. Отсутствующие Фигурные Скобки

Отсутствие фигурных скобок может привести к ошибке “незаконное начало выражения”. Давайте сначала рассмотрим пример:

package com.baeldung;

public class MissingCurlyBraces {
    public void printSum(int x, int y) {
        System.out.println("Calculation Result:" + calcSum(x, y));
        
    public int calcSum(int x, int y) {
        return x + y;
    }
}

Если мы скомпилируем вышеуказанный класс:

$ javac MissingCurlyBraces.java
MissingCurlyBraces.java:7: error: illegal start of expression
        public int calcSum(int x, int y) {
        ^
MissingCurlyBraces.java:7: error: ';' expected
        public int calcSum(int x, int y) {
   .....

Отсутствие закрывающей фигурной скобки print Sum() является основной причиной проблемы.

Решение проблемы простое — добавление закрывающей фигурной скобки в метод printSum() :

package com.baeldung;

public class MissingCurlyBraces {
    public void printSum(int x, int y) {
        System.out.println("Calculation Result:" + calcSum(x, y));
    }
    public int calcSum(int x, int y) {
        return x + y;
    }
}

Прежде чем перейти к следующему разделу, давайте рассмотрим ошибку компилятора.

Компилятор сообщает, что 7-я строка вызывает ошибку “незаконное начало выражения”. На самом деле, мы знаем, что первопричина проблемы находится в 6-й строке. Из этого примера мы узнаем, что иногда ошибки компилятора не указывают на строку с основной причиной , и нам нужно будет исправить синтаксис в предыдущей строке.

3. Модификатор Доступа Внутри Метода

В Java мы можем объявлять локальные переменные только внутри метода или конструктора . Мы не можем использовать модификатор доступа для локальных переменных внутри метода, поскольку их доступность определяется областью действия метода.

Если мы нарушим правило и у нас будут модификаторы доступа внутри метода, возникнет ошибка “незаконное начало выражения”.

Давайте посмотрим на это в действии:

package com.baeldung;

public class AccessModifierInMethod {
    public void printSum(int x, int y) {
        private int sum = x + y; 
        System.out.println("Calculation Result:" + sum);
    }
}

Если мы попытаемся скомпилировать приведенный выше код, мы увидим ошибку компиляции:

$ javac AccessModifierInMethod.java 
AccessModifierInMethod.java:5: error: illegal start of expression
        private int sum = x + y;
        ^
1 error

Удаление модификатора private access легко решает проблему:

package com.baeldung;

public class AccessModifierInMethod {
    public void printSum(int x, int y) {
        int sum = x + y;
        System.out.println("Calculation Result:" + sum);
    }
}

4. Вложенные методы

Некоторые языки программирования, такие как Python, поддерживают вложенные методы. Но, Java не поддерживает метод внутри другого метода.

Мы столкнемся с ошибкой компилятора “незаконное начало выражения”, если создадим вложенные методы:

package com.baeldung;

public class NestedMethod {
    public void printSum(int x, int y) {
        System.out.println("Calculation Result:" + calcSum(x, y));
        public int calcSum ( int x, int y) {
            return x + y;
        }
    }
}

Давайте скомпилируем приведенный выше исходный файл и посмотрим, что сообщает компилятор Java:

$ javac NestedMethod.java
NestedMethod.java:6: error: illegal start of expression
        public int calcSum ( int x, int y) {
        ^
NestedMethod.java:6: error: ';' expected
        public int calcSum ( int x, int y) {
                          ^
NestedMethod.java:6: error:  expected
        public int calcSum ( int x, int y) {
                                   ^
NestedMethod.java:6: error: not a statement
        public int calcSum ( int x, int y) {
                                        ^
NestedMethod.java:6: error: ';' expected
        public int calcSum ( int x, int y) {
                                         ^
5 errors

Компилятор Java сообщает о пяти ошибках компиляции. В некоторых случаях одна ошибка может привести к нескольким дальнейшим ошибкам во время компиляции.

Выявление первопричины имеет важное значение для того, чтобы мы могли решить эту проблему. В этом примере первопричиной является первая ошибка “незаконное начало выражения”.

Мы можем быстро решить эту проблему, переместив метод calcSum() из метода print Sum() :

package com.baeldung;

public class NestedMethod {
    public void printSum(int x, int y) {
        System.out.println("Calculation Result:" + calcSum(x, y));
    }
    public int calcSum ( int x, int y) {
        return x + y;
    }
}

5. символ или строка Без кавычек

Мы знаем, что String литералы должны быть заключены в двойные кавычки, в то время как char значения должны быть заключены в одинарные кавычки.

Если мы забудем заключить их в соответствующие кавычки, компилятор Java будет рассматривать их как имена переменных .

Мы можем увидеть ошибку “не удается найти символ”, если “переменная” не объявлена.

Однако если мы забудем дважды заключить в кавычки Строку , которая не является допустимым именем переменной Java , компилятор Java сообщит об ошибке “незаконное начало выражения” .

Давайте посмотрим на это на примере:

package com.baeldung;

public class ForgetQuoting {
    public int calcSumOnly(int x, int y, String operation) {
        if (operation.equals(+)) {
            return x + y;
        }
        throw new UnsupportedOperationException("operation is not supported:" + operation);
    }
}

Мы забыли процитировать строку |//+ внутри вызова метода equals , и + , очевидно, не является допустимым именем переменной Java.

Теперь давайте попробуем его скомпилировать:

$ javac ForgetQuoting.java 
ForgetQuoting.java:5: error: illegal start of expression
        if (operation.equals(+)) {
                              ^
1 error

Решение проблемы простое — обертывание String литералов в двойные кавычки:

package com.baeldung;

public class ForgetQuoting {
    public int calcSumOnly(int x, int y, String operation) {
        if (operation.equals("+")) {
            return x + y;
        }
        throw new UnsupportedOperationException("operation is not supported:" + operation);
    }
}

6. Заключение

В этой короткой статье мы рассказали о пяти различных сценариях, которые приведут к ошибке “незаконное начало выражения”.

В большинстве случаев при разработке приложений Java мы будем использовать среду IDE, которая предупреждает нас об обнаружении ошибок. Эти замечательные функции IDE могут значительно защитить нас от этой ошибки.

Тем не менее, мы все еще можем время от времени сталкиваться с этой ошибкой. Поэтому хорошее понимание ошибки поможет нам быстро найти и исправить ошибку.