Else without if ошибка java

Java Hungry

Java developers tutorials and coding.

[Solved] else without if error

else without if error is mostly faced by the java beginners. As the name suggests java compiler is unable to find an if statement associated with your else statement. else statements do not execute unless they are associated with the if statement. As always, first, we will produce the error before moving on to the solution.

[Fixed] error: ‘else’ without ‘if’

Example 1: Producing the error by ending if statement with a semi-colon

Output:
/ElseWithoutIf.java:8: error: ‘else’ without ‘if’
else <
^
1 error

Explanation:

The statement if(input > 100); is equivalent to writing
if (input > 100)
<
>
i.e. input is greater than 100 then execute an empty block. There is no else block for this. As a result, the next line else block does not have a corresponding if block, thus, produces the error: else without if.

Solution:

The above compilation error can be resolved by removing the semicolon.

Output:
input is less than or equal to 100

Example 2: Producing the error by missing the closing bracket in if condition

Just like the above example, we will produce the error first before moving on to the solution.

Explanation:

In the above example, if condition closing bracket is missing, hence, we are getting the else without if error.

Solution:

The above compilation error can be resolved by inserting the missing closing bracket as shown below.

Output:
inside else

Example 3: Producing the error by having more than one else clause for an if statement

Output:
/ElseWithoutIf3.java:12: error: ‘else’ without ‘if’
else
^
1 error

Explanation:

In the above scenario, you are chaining the two elses together that is not correct syntax, hence, resulting in the error. The correct syntax is given below:

Solution:

The above compilation error can be resolved by providing correct if-elseif-else syntax as shown below.

Output:
input is less than 100

The best way to deal with error: ‘else’ without ‘if’ is always to use curly braces to show what is executed after if. Learn to indent the code too as it helps in reading and understanding.

Источник

Understanding Common Errors In Java

When we were building our automated common error feedback feature on Mimir Classroom, we analyzed millions of stack traces that our students received when completing their coursework on our platform. In this post, you will find the errors we found to be most troubling to students in Java and some tips on how to approach them.

To start off, here is a quick refresher on how to read an error in the Java stack trace.

1. ’;’ expected

This error means that you forgot a semicolon in your code. If you look at the full error statement, it will tell you where in your code you should check within a few lines. Remember that all statements in Java must end in a semicolon and elements of Java like loops and conditionals are not considered statements.

2. cannot find symbol

This error means that Java is unable to find an identifier that it is trying to work with. It is most often caused by the following:

• You misspelled a variable name, class name, or a keyword somewhere. Remember that Java is case sensitive.
• You are trying to access a variable or class which does not exist or can not be accessed.
• You forgot to import the right class.

3. illegal start of expression

This error usually occurs when your code does not follow Java’s standard structure. Check the nesting of your variables and ensure that all your < >and ( ) are in the right place..

4. class, interface, or enum expected

This error is usually caused by misplaced < >. All code in Java needs to be contained within a class, interface, or enum. Ensure that all of your code is within the < >of one of those. We often have seen this error when there is an extra > at the end of your code.

5. reached end of file while parsing

This error usually occurs when you are missing a > somewhere in your code. Ensure that for every < you have one >in the right place.

6. missing return statement

This error usually means one of two things:

• Your method is expecting a return statement but is missing one. Ensure that if you declared a method that returns something other than void and that it returns the proper variable type.
• Your return statements are inside conditionals who’s parameters may not be reached. In this case, you will need to add an additional return statement or modify your conditionals.

7. ‘else’ without ‘if’

This error means that Java is unable to find an if statement associated to your else statement. Else statements do not work unless they are associated with an if statement. Ensure that you have an if statement and that your else statement isn’t nested within your if statement.

8. ‘(‘ expected or ‘)’ expected

This error means that you forgot a left or right parenthesis in your code. If you look at the full error statement, it will tell you where in your code you should check. Remember that for every ( you need one ).

9. case, default, or ‘>’ expected

This error means that your switch statement isn’t properly structured. Ensure that your switch statement contains a variable like this: switch (variable). Also ensure that every case is followed by a colon (:) before defining your case.

10. ‘.class’ expected

This error usually means that you are trying to declare or specify a variable type inside of return statement or inside of a method calls. For example: “return int 7;» should be «return 7;»

11. invalid method declaration; return type required

Every Java method requires that you declare what you return even if it is nothing. «public getNothing()» and «public static getNumber()» are both incorrect ways to declare a method.

The correct way to declare these is the following: «public void getNothing()» and «public static int getNumber()»

12. unclosed character literal

This error occurs when you start a character with a single quote mark but don’t close with a single quote mark.

13. unclosed string literal

This error occurs when you start a string with a quotation mark but don’t close with a second quotation mark. This error can also occur when quotation marks inside strings are not escaped with a backslash.

14. incompatible types

This error occurs when you use the wrong variable type in an expression. A very common example of this is sending a method an integer when it is expecting a string. To solve this issue, check the types of your variables and how they are interacting with other types. There are also ways to convert variable types.

15. missing method body

This error commonly occurs when you have a semicolon on your method declaration line. For example «public static void main(String args[]);». You should not have a semicolon there. Ensure that your methods have a body.

16. unreachable statement

This error means that you have code that will never be executed. Usually, this is after a break or a return statement.

Источник

1. the error: 'else' without 'if' in Java
2. Reasons for error: 'else' without 'if' in Java
3. Fix the error: 'else' without 'if' in Java

Today, we will learn about an error saying 'else' without 'if' while writing code in Java. We will also figure out the possible reasons causing this error and find its solution.

the error: 'else' without 'if' in Java

Usually, this kind of error is faced by newbies in Java programming. Before moving toward the causes and solution for this error, let’s write a program that produces this error and understand it.

So, assuming that we are Python experts and beginners in Java. So, we will write the Java program containing if-else as follows.

Example Code:

//import libraries
import java.util.Scanner;

//decide future activity based on the current temperature
public class Test{
    public static void main (String[] args){

        int temp;
        Scanner scan = new Scanner(System.in);
        System.out.println ("What's the current temperature?");
        temp = scan.nextInt();

        if (temp > 95 || temp < 20);
            System.out.println ("Visit our shops");
            else if (temp <= 95)
                if (temp >= 80)
                System.out.println ("Swimming");
                else if (temp >=60)
                    if (temp <= 80)
                    System.out.println ("Tennis");
                    else if (temp >= 40)
                        if (temp < 60)
                        System.out.println ("Golf");
                        else if (temp < 40)
                            if (temp >= 20)
                            System.out.println ("Sking");                                      }//end main()
}//end Test Class

Error:

In this program, we get the current temperature from the user and decide our future activity based on the current temperature. The image above shows that we are getting a logical error about which NetBeans IDE informs at compile time.

So, we cannot even execute the code until we resolve the error. For that, we will have to know the possible reasons below.

Reasons for error: 'else' without 'if' in Java

The error itself is explanatory, which says that a Java compiler cannot find an if statement associated with the else statement. Remember that the else statement does not execute unless they’re associated with an if statement.

So, the possible reasons are listed below.

1. The first reason is that we forgot to write the if block before the else block.
2. The closing bracket of the if condition is missing.
3. We end the if statement by using a semi-colon.

How to solve this error? Let’s have a look at the following section.

Fix the error: 'else' without 'if' in Java

Example Code:

//import libraries
import java.util.Scanner;

public class Test {
    public static void main(String[] args) {

        int temp;
        Scanner scan = new Scanner(System.in);
        System.out.println("What's the current temperature?");
        temp = scan.nextInt();

        if (temp > 95 || temp < 20) {
            System.out.println("Visit our shops");
        }//end if
        else if (temp <= 95) {
            if (temp >= 80) {
                System.out.println("Swimming");
            } //end if
            else if (temp >= 60) {
                if (temp <= 80) {
                    System.out.println("Tennis");
                }//end if
                else if (temp >= 40) {
                    if (temp < 60) {
                        System.out.println("Golf");
                    }//end if
                    else if (temp < 40) {
                        if (temp >= 20) {
                            System.out.println("Sking");
                        }//end if
                    }//end else-if
                }//end else-if
            }//end else-if
        }//end else-if
    }//end main()
}//end Test Class

Output:

What's the current temperature?
96
Visit our shops

We removed the semi-colon (;) from the end of the if statement and placed the {} for each block to fix an error saying 'else' without 'if'.

It is better to use curly brackets {} until we are expert enough and know where we can omit them (we can omit them when we have a single statement in the block).

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.