Ошибка компилятора Java: незаконное начало выражения
См. Примеры, иллюстрирующие основные причины ошибки “незаконное начало выражения” и способы ее устранения.
- Автор записи
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:
Если мы скомпилируем вышеуказанный класс:
Отсутствие закрывающей фигурной скобки print Sum() является основной причиной проблемы.
Решение проблемы простое — добавление закрывающей фигурной скобки в метод printSum() :
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:
Если мы попытаемся скомпилировать приведенный выше код, мы увидим ошибку компиляции:
Удаление модификатора private access легко решает проблему:
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:
Давайте скомпилируем приведенный выше исходный файл и посмотрим, что сообщает компилятор Java:
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() :
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:
Мы забыли процитировать строку |//+ внутри вызова метода equals , и + , очевидно, не является допустимым именем переменной Java.
Теперь давайте попробуем его скомпилировать:
Решение проблемы простое — обертывание String литералов в двойные кавычки:
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.
How to Fix “Illegal Start of Expression” in Java
Over the past two and a half decades, Java has consistently been ranked as one of the top 3 most popular programming languages in the world [1], [2]. As a compiled language, any source code written in Java needs to be translated (i.e., compiled) into machine code before it can be executed. Unlike other compiled languages where programs are compiled directly into machine code, the Java compiler converts the source code into intermediate code, or bytecode, which is then translated into machine code for a specific platform by the Java Virtual Machine (JVM). This, in the simplest of terms, is how Java achieves its platform independence (Fig. 1).
One advantage that comes with being a compiled language is the fact that many errors stemming from incorrect language syntax and semantics (such as “illegal start of expression”) can be captured in the compilation process, before a program is run and they inadvertently find their way into production environments. Since they occur at the time of compilation, these errors are commonly referred to as compile-time errors.
The Java compiler can detect syntax and static semantic errors, although it is incapable of recognizing dynamic semantic errors. The latter are logical errors that don’t violate any formal rules and as such cannot be detected at compile-time; they only become visible at runtime and can be captured by well-designed tests.
When it encounters an error it can recognize, the Java compiler generates a message indicating the type of error and the position in the source file where this error occurred. Syntax errors are the easiest to detect and correct.
Figure 1: The Java Compilation Process [3]
Illegal Start of Expression: What is it?
Expressions are one of the main building blocks of any Java application. These are constructs that compute values and control the execution flow of the program. As its name implies, the “illegal start of expression” error refers to an expression that violates some rule at the point where it starts, usually right after another expression ends; the assumption here is that the preceding expression is correct, i.e., free of errors.
The “illegal start of expression” error often arises from an insufficient familiarity with the language or due to basic negligence. The cause for this error can usually be found at the beginning of an expression or, in some cases, the entire expression might be incorrect or misplaced.
Illegal Start of Expression Examples
Access modifiers on local variables
A local variable in Java is any variable declared inside the body of a method or, more generally, inside a block. A local variable’s accessibility is predetermined by the block in which it is declared—the variable can be accessed strictly within the scope of its enclosing block. Therefore, access modifiers have no use here and, if introduced, will raise the “illegal start of expression” error (Fig. 2(a)). Removing the access modifier (as shown on line 5 in Fig. 2(b)) resolves the Java error.
Figure 2: Local variable with an access modifier (a) error and (b) resolution
Nested methods
Unlike some other languages (most notably functional languages), Java does not allow direct nesting of methods, as shown in Fig. 3(a). This violates Java’s scoping rules and object-oriented approach.
There are two main ways of addressing this issue. One is to move the inner method to an appropriate place outside the outer method (Fig. 3(b)). Another one is to replace the inner method with a lambda expression assigned to a functional interface (Fig. 3(c)).
Figure 3: Nested method (a) error and (b)(c) two viable resolutions
Missing braces
According to Java syntax, every block has to start and end with an opening and a closing curly brace, respectively. If a brace is omitted, the compiler won’t be able to identify the start and/or the end of a block, which will result in an illegal start of expression error (Fig. 4(a)). Adding the missing brace fixes the error (Fig. 4(b)).
Figure 4: Missing curly brace (a) error and (b) resolution
Array creation
Traditionally, array creation in Java is done in multiple steps, where the array data-type and size are declared upfront and its values initialized afterwards, by accessing its indices. However, Java allows doing all of these operations at once with a succinct, albeit somewhat irregular-looking, syntax (Fig. 5(a)).
While very convenient, this syntactical idiosyncrasy only works as a complete inline expression and will raise the illegal start of expression error if used otherwise (Fig. 5(b)). This syntax cannot be used to initialize values of an array whose size has already been defined, because one of the things it tries to do is exactly that—assign a size to the array.
The only other scenario in which this syntax may be used is to overwrite an existing array with a new one, by prefixing it with the new directive (Fig. 5(c)).
Figure 5: Array creation (a)(c) valid syntax and (b) invalid syntax examples
Summary
Being a compiled language, Java has an advantage over other languages in its ability to detect and prevent certain errors from slipping through into production. One such error is the “illegal start of expression” error which belongs to the category of syntax errors detected at compile time. Common examples have been presented in this article along with explanations of their cause and ways to resolve them.
Track, Analyze and Manage Errors With Rollbar
Managing errors and exceptions in your code is challenging. It can make deploying production code an unnerving experience. Being able to track, analyze, and manage errors in real-time can help you to proceed with more confidence. Rollbar automates error monitoring and triaging, making fixing Java errors easier than ever. Sign Up Today!
References
[1] TIOBE Software BV, “TIOBE Index for October 2021: TIOBE Programming Community index,” TIOBE Software BV. [Online]. Available: https://www.tiobe.com/tiobe-index/. [Accessed Oct. 28, 2021].
[2] Statistics & Data, “The Most Popular Programming Languages – 1965/2021,” Statistics and Data. [Online]. Available: https://statisticsanddata.org/data/the-most-popular-programming-languages-1965-2021/. [Accessed Oct. 28, 2021].
[3] C. Saternos, Client-Server Web Apps with JavaScript and Java. Sebastopol, CA: O’Reilly Media, Inc., 2014, Ch. 4, p.59
Ошибка компиляции: illegal start of expression
В методе Goperation() — scc.hasNext() — возращает boolean — правда/ не правда, т.е. вы не записываете чар, а проверяете наличие. «Существует и метод hasNext() , проверяющий остались ли в потоке ввода какие-то символы.» http://kostin.ws/java/java-input-stream.html
- Если вы используете char — то нужно указывать » , вместо «» — они используются для String .
(9:9)illegal start of expression. — говорит что это ошибка компиляции. Структура программы на Java имеет определённый синтаксис. Этот синтаксис определяет правила использования элементов языка в вашей программе.
Java это очень сложный язык. Конечно читать сразу JLS будет сложно, поэтому новичкам рекомендуют начинить с пониманием базовых концепций. Потом уже переходить к кодированию.
How to fix «illegal start of expression» error in Java? Example
When the compiler checks the next statement it sees an illegal start because an earlier statement was not terminated. The bad part is that you can get tens of «illegal start of expression» errors by just omitting a single semi-colon or missing braces, as shown in the following example.
If you compile this program you will be greeted with several «illegal start of expression» error as shown below:
$ javac Main.java
Main.java:7: error: illegal start of expression
public static int count() <
^
Main.java:7: error: illegal start of expression
public static int count() <
^
Main.java:7: error: ‘;’ expected
public static int count() <
^
Main.java:7: error: ‘;’ expected
public static int count() <
^
Main.java:11: error: reached end of file while parsing
>
^
5 errors
And the only problem is missing curly braces in the main method, as soon as you add that missing closing curly brace in the main method the code will work fine and all these errors will go away, as shown below:
The error message is very interesting, if you look at the first one, you will find that error is in line 7 and the compiler complaining about a public keyword but the actual problem was in line 5 where a closing brace is missing.
From the compiler’s view, the public should not come there because it’s an invalid keyword inside a method in Java, remember without closing braces main() is still not closed, so the compiler thinks the public keyword is part of the main() method.
Though, when you compile the same program in Eclipse, you get a different error because its the compiler is slightly different than javac but the error message is more helpful as you can see below:
Exception in thread «main» java.lang.Error: Unresolved compilation problem:
Syntax error, insert «>» to complete MethodBody
In short, the «illegal start of expression» error means the compiler find something inappropriate, against the rules of Java programming but the error message is not very helpful. For «illegal start of expression» errors, try looking at the lines preceding the error for a missing ‘)’ or ‘>’ or missing semicolon.
Also remember, a single syntax error somewhere can cause multiple «illegal start of expression» errors. Once you fix the root cause, all errors will go away. This means always recompile once you fix an error, don’t try to make multiple changes without compilation.
- How to fix «variable might not have been initialized» error in Java? (solution)
- Could not create the Java virtual machine Invalid maximum heap size: -Xmx (solution)
- How to fix class, enum, or interface expected error in Java? (solution)
- How to fix java.lang.ClassNotFoundException: org.apache.commons.logging.LogFactory (solution)
- Caused By: java.lang.NoClassDefFoundError: org/apache/log4j/Logger in Java (solution)
- Error: could not open ‘C:\Java\jre8\lib\amd64\jvm.cfg’ (solution)
- java.lang.OutOfMemoryError: Java heap space : Cause and Solution (steps)
- How to avoid ConcurrentModificationException while looping over List in Java? (solution)
78 comments:
thanks this really helped alot
Hello @Unknown, can you provide more details so that we can help you better?
import java.lang.*;
import java.io.*;
import java.util.*;
class keyboard
<
public static void main(String s[] )throws IOException
<
int a[]= Integer.parseInt(a[10]);
int b[]=Integer.parseInt(b[10]);
int c[]=Integer.parseInt(c[10]);
InputStreamReader obj1= new InputStreamReader(System.in);
BufferedReader obj2=new BufferedReader(obj1);
System.out.println(a);
System.out.println(b);
System.out.println(c);
>
>
System.out.println("I'm fine.";) What I'm I missing
it should look like (I'm fine). on the display
your semicolon is inside the parentheses, and if you want to see the parentheses you will have to put it inside the " " cause it only prints whats inside those
What if the main problem is the public static void main(string args)
here is my program please tell what is the mistake.
couple of mistakes, first 1. class is not close with ">" which is missing
2. syntax of main method is not correct, if you really intent to use main() method
// whats the mistakes?
// Java program to Draw a
// Smiley using Java Applet
import java.applet.*;
import java.awt.*;
import java.util.*;
class Rextester
<
public static void main(string[] args)
<
public class Rextester extends Applet <
public void paint(Graphics g)
<
// Oval for face outline
g.drawOval(80, 70, 150, 150);
// Ovals for eyes
// with black color filled
g.setColor(Color.BLACK);
g.fillOval( 120, 120, 15, 15);
g.fillOval(170, 120, 15, 15);
// Arc for the smile
g.drawArc(130,180,50,20,180,180);
>
>
>
>
you are declaring a class inside a static method, that cannot be public.
S would be capital in main(String args[])
Hi, may I know what is the mistakes?
public class Recognizer <
public static void main(String[] args)
<
private static final int MAXN=10;
private static final int MAXNUM=9;
private static final int[][][][] count=new int[MAXN][MAXN][2][MAXNUM+1];
private static final int[] count2=new int[MAXNUM+1];
private static final double[][][][] po=new double[MAXN][MAXN][2][MAXNUM+1];
private static final double[] po2=new double[MAXNUM+1];
public static double[] recognize(boolean[][] input)
<
double[] result=new double[MAXNUM+1];
for (int i=0;i<=MAXNUM;i++)
<
result[i]=1;
for (int j=0;j<MAXN;j++)
for (int k=0;k<MAXN;k++)
if (input[j][k])
result[i]*=po[j][k][1][i];
else
result[i]*=po[j][k][0][i];
result[i]*=po2[i];
>
return result;
>
>
>
Guys can you help me? What the wrong mistake in my code 🙁
public class BubbleSort <
public String version()
< return "1.2.02";>
static void BubbleSort(int[] angka) <
for (int i = 0; i < angka.length; i++) <
for (int j = 0; j < angka.length-1; j++) <
if (angka[j]>angka[j+1]) <
int temp = angka [j];
angka [j] = angka[j+1];
angka[j+1] = temp;
for (int i = 0; i < angka.length; i++) <
System.out.print(angka[i]+" ");
>
>;
System.out.println("");
BubbleSort(angka);
for (int i = 0; i < angka.length; i++) <
System.out.print(angka[i]+" ");
>);
>
>
what is the error you are getting?
import java.io.*;
import java.util.*;
class calci<
int a,b,c,d,e,f,g;<
Scanner cal= new Scanner(System.in);
System.out.println("Enter the first number :");
a = cal.nextInt();
System.out.println("Enter the second number :");
b = cal.nextInt();
public static int add()
<
c=a+b;
System.out.println("Addition of 2 numbers is"+c);
return c;
>
public static void sub()
<
d=a-b;
System.out.println("Subtraction of 2 numbers is ="+d);
return d;
>
public static int mul()
<
e=a*b;
System.out.println("Multiplication of 2 numbers is ="+e);
return e;
>
public static int div()
<
f=a/b;
System.out.println("Division of 2 numbers is ="+f);
return f;
>
public static int mod()
<
g=a%b;
System.out.println("Modulo of 2 numbers is ="+g);
return g;
>
>
>
class calculator_2<
public static void main(String [] args)<
int ch;
System.out.println("1.ADDITION \n2.SUBTRACTION \n3.MULTIPLICATION \n4.DIVISION \n5.MODULO");
calci obj=new calci();
Scanner in =new Scanner(System.in);
System.out.println("Enter Your Choice :\t");
ch = in.nextInt();
switch(ch)
<
case 1: obj.add();
break;
case 2: bj.sub();
break;
case 3: obj. mul();
break;
case 4: obj.div();
break;
case 5: obj.mod();
break;
default :System.out.println("option not avilable");
break;
>
>
>
My program is showing error in public static void
can you post your program here?
Poor Javin got flooded by help requests.
1
public class temperature
2
<
3
public static void main(String []args)
4
<
5
double tempC,tempF;
6
if(choice==0)
7
<
8
tempC = 37;
9
tempF = (tempC*1.8F)+32;
10
System.out.println("temperature in farenheit is " +tempF);
11
>
12
else if (choice==2)
13
<
14
tempF = 98.4;
15
tempC = (tempF-32)/1.8F;
16
System.out.println("farenheit in temperature is " +tempC);
17
>
18
else
19
System.out.println("enter 1 for-C-to -F and 2 for F-to -C conversion");
20
>
21
>
22
23
This code is editable. Click Run to Compile + Execute
Line 7: error: illegal start of expression
if(choice = =0 )
^
Line 13: error: illegal start of expression
else if (choice = = 2)
^
2 errors
what are the mistakess in this?
because you have not declared the "choice" variable before using it. Just declare choice at the first line of main method like int choice = 0 it will work.
like
int choice=0;
double tempC, tempF;
import java.util.Scanner;
class salpha <
public static void main(String args[]) <
Scanner b=new Scanner(System.in);
System.out.println("enter size");
int n;
n=b.nextInt();
for(int i=0;i<=n;i++) <
for(int j=0;j<=n;j++) <
if(( i==0||i==n)&& i!=0 ||i==0 &&j!=0 && j!=n||)
IT WAS SHOWING MISTAKE AT IF ILLEGAL START OF EXPRESSION
You can also get this for the -> Symbol in removeIf.. When using Java version < 8.
absolutely true, and the judging system of my university uses Java 1.8.
public static void main(String [] args) <
System.out.println("your art goes here");
System.out.println("across many lines");
>
. ________ .
System.out.println(" /\\ / \\ \\\\ \\ / \\");
System.out.println(" / \\ / . \\ _____ \\ ____ \\ / . \\");
System.out.println("/ | \\ \\ / / \\___\\ / / | | \\ | / / \\___\");
System.out.println("| | \\ | | | / / \\ | | | | | |");
System.out.println("| | \\ | | | / /_____ | | | | | |");
System.out.println(" | |____| | \\ | | / / | | | | | |");
System.out.println(" | ______ | | | /_____ / | | | | | |");
System.out.println(" | | | | | | / / | | | | | |");
System.out.println("| | | | | | ___ / / | | / | | | "); __");
System.out.println("| | | | \\ \\./ / / / | |__/ | \\ \\./ /");
System.out.println("/ \\ / \\ \\ / / / / / \\ /");
System.out.println("\\./ \\./ \\./ / /________/ \\./ ");>
can i know he error in that program
I think it's because you ended the code after the first two println's you need to move your curly brace to the end of the program for it to work.
/* Leap year __________
____________ Created by Sourav Dutta*/
public class Program
<
public static void main(String[] args) <
public static boolean isLeapYear(int year) <
year=1600;
if (year % 4 != 0) <
return false;
>
else if (year % 400 == 0) <
return true;
>
else if (year % 100 == 0) <
return false;
>
else <
return true;>
Nice one Sourav, but you can optimize the logic by using switch statement, it would be much cleaner to read.
Can you help me on this program
import java.util.Scanner;
public class DemoVariables
<
int entry;
Scanner keyBoards = new Scanner(System.in);
System.out.print("Enter an integer");
entry = keyBoard.nextInt();
public static void main(String[] args)
<
System.out.print("The other entry is");
System.out.println("anotherEntry");
System.out.println(entry + " plus " +
anotherEntry + " is " + ( entry + anotherEntry));
System.out.println(entry + " minus " +
anotherEntry + " is " + (entry — anotherEntry));
System.out.println(entry + " times " +
anotherEntry + " is +(entry * anotherEntry));
System.out.println(entry + " divided by " +
anotherEntry + " is " + (entry / anotherEntry));
System.out.println("The remainder is" +
(entry % anotherEntry));
>
>
Code:
import java.io.*;
class SwitchMethod<
public static void main(String[] ar) throws IOException<
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter a Choice of Your Number From 1 to 4 :");
int ch = Integer.parseInt(br.readLine());
SwitchMethod sm= new SwitchMethod();
switch(ch)<
case 1:
void add() throws IOException<
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.print("Enter First Value:");
int x = Integer.parseInt(br.readLine());
System.out.print("Enter Second Value:");
int y = Integer.parseInt(br.readLine());
int z = x+y;
System.out.println(" Sum of " +x+ " and " +y+ " is: " +z);
sm.add();
>
break;
case 2: sm.sub();
break;
int sub() throws IOException <
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter First Value:");
int a = Integer.parseInt(br.readLine());
System.out.println("Enter Second Value:");
int b = Integer.parseInt(br.readLine());
int c = a-b;
return c;
int res = sm.sub();
System.out.println("Sum:"+res);
>
case 3: sm.mul(a,b);
break;
void mul(int x,int y) throws IOException <
int z = x*y;
System.out.println("Multiplying " +x+ "*" +y+ "is:" +z);
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println("Enter First Value:");
int a = Integer.parseInt(br.readLine());
System.out.println("Enter Second Value:");
int b = Integer.parseInt(br.readLine()); mp.mul(a,b);
>
case 4: int div(int x,int y) throws IOException <
int z = x/y;
return z;
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.print("Enter First Value:");
int a = Integer.parseInt(br.readLine());
System.out.print("Enter Second Value:");
int b = Integer.parseInt(br.readLine());
int res = sm.div(a,b);
System.out.println("Sum: " +res);
>
break;
Command Prompt:
SwitchMethod.java:10: error: illegal start of expression
void add() throws IOException<
^
SwitchMethod.java:10: error: ';' expected
void add() throws IOException <
^
SwitchMethod.java:10: error: not a statement
void add() throws IOException <
^
SwitchMethod.java:10: error: ';' expected
void add() throws IOException <
^
SwitchMethod.java:23: error: ';' expected
int sub() throws IOException <
^
SwitchMethod.java:23: error: not a statement
int sub() throws IOException <
^
SwitchMethod.java:23: error: ';' expected
int sub() throws IOException <
^
SwitchMethod.java:36: error: illegal start of expression
void mul(int x,int y) throws IOException <
^
SwitchMethod.java:36: error: ';' expected
void mul(int x,int y) throws IOException <
^
SwitchMethod.java:36: error: expected
void mul(int x,int y) throws IOException <
^
SwitchMethod.java:36: error: ';' expected
void mul(int x,int y) throws IOException <
^
SwitchMethod.java:36: error: not a statement
void mul(int x,int y) throws IOException <
^
SwitchMethod.java:36: error: ';' expected
void mul(int x,int y) throws IOException <
^
SwitchMethod.java:45: error: ';' expected
case 4: int div(int x,int y) throws IOException <
^
SwitchMethod.java:45: error: expected
case 4: int div(int x,int y) throws IOException <
^
SwitchMethod.java:45: error: ';' expected
case 4: int div(int x,int y) throws IOException <
^
SwitchMethod.java:45: error: not a statement
case 4: int div(int x,int y) throws IOException <
^
SwitchMethod.java:45: error: ';' expected
case 4: int div(int x,int y) throws IOException <
^
18 errors
just copy paste your code into an IDE like Eclipse and it will tell you what is wrong with that instead of giving so many errors.