Java application error illegalargumentexception

An unexpected event occurring during the program execution is called an Exception. This can be caused due to several factors like invalid user input, network failure, memory limitations, trying to open a file that does not exist, etc. 

If an exception occurs, an Exception object is generated, containing the Exception’s whereabouts, name, and type. This must be handled by the program. If not handled, it gets past to the default Exception handler, resulting in an abnormal termination of the program.

IllegalArgumentException 

The IllegalArgumentException is a subclass of java.lang.RuntimeException. RuntimeException, as the name suggests, occurs when the program is running. Hence, it is not checked at compile-time.

IllegalArgumentException Cause

When a method is passed illegal or unsuitable arguments, an IllegalArgumentException is thrown.

The program below has a separate thread that takes a pause and then tries to print a sentence. This pause is achieved using the sleep method that accepts the pause time in milliseconds. Java clearly defines that this time must be non-negative. Let us see the result of passing in a negative value.

Program to Demonstrate IllegalArgumentException:

Output:

Exception in thread "Test Thread" java.lang.IllegalArgumentException: 
timeout value is negative
    at java.base/java.lang.Thread.sleep(Native Method)
    at Main$1.run(Main.java:19)
    at java.base/java.lang.Thread.run(Thread.java:834)

In the above case, the Exception was not caught. Hence, the program terminated abruptly and the Stack Trace that was generated got printed.

Diagnostics & Solution

The Stack Trace is the ultimate resource for investigating the root cause of Exception issues. The above Stack Trace can be broken down as follows.

Part 1: This part names the Thread in which the Exception occurred. In our case, the Exception occurred in the “Test Thread”.

Part 2: This part names class of the Exception. An Exception object of the “java.lang.IllegalArgumentException” class is made in the above example.

Part 3: This part states the reason behind the occurrence of the Exception. In the above example, the Exception occurred because an illegal negative timeout value was used.

Part 4: This part lists all the method invocations leading up to the Exception occurrence, beginning with the method where the Exception first occurred. In the above example, the Exception first occurred at Thread.sleep() method. 

From the above analysis, we reach the conclusion that an IllegalArgumentException occurred at the Thread.sleep() method because it was passed a negative timeout value. This information is sufficient for resolving the issue. Let us accordingly make changes in the above code and pass a positive timeout value.

Below is the implementation of the problem statement:

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How to Solve IllegalArgumentException in Java?

An unexpected event occurring during the program execution is called an Exception. This can be caused due to several factors like invalid user input, network failure, memory limitations, trying to open a file that does not exist, etc.

If an exception occurs, an Exception object is generated, containing the Exception’s whereabouts, name, and type. This must be handled by the program. If not handled, it gets past to the default Exception handler, resulting in an abnormal termination of the program.

IllegalArgumentException

The IllegalArgumentException is a subclass of java.lang.RuntimeException. RuntimeException, as the name suggests, occurs when the program is running. Hence, it is not checked at compile-time.

IllegalArgumentException Cause

When a method is passed illegal or unsuitable arguments, an IllegalArgumentException is thrown.

The program below has a separate thread that takes a pause and then tries to print a sentence. This pause is achieved using the sleep method that accepts the pause time in milliseconds. Java clearly defines that this time must be non-negative. Let us see the result of passing in a negative value.

Program to Demonstrate IllegalArgumentException:

Output:

In the above case, the Exception was not caught. Hence, the program terminated abruptly and the Stack Trace that was generated got printed.

Diagnostics & Solution

The Stack Trace is the ultimate resource for investigating the root cause of Exception issues. The above Stack Trace can be broken down as follows.

Part 1: This part names the Thread in which the Exception occurred. In our case, the Exception occurred in the “Test Thread”.

Part 2: This part names class of the Exception. An Exception object of the “java.lang.IllegalArgumentException” class is made in the above example.

Part 3: This part states the reason behind the occurrence of the Exception. In the above example, the Exception occurred because an illegal negative timeout value was used.

Part 4: This part lists all the method invocations leading up to the Exception occurrence, beginning with the method where the Exception first occurred. In the above example, the Exception first occurred at Thread.sleep() method.

From the above analysis, we reach the conclusion that an IllegalArgumentException occurred at the Thread.sleep() method because it was passed a negative timeout value. This information is sufficient for resolving the issue. Let us accordingly make changes in the above code and pass a positive timeout value.

Below is the implementation of the problem statement:

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How to Throw IllegalArgumentException in Java

Table of Contents

The IllegalArgumentException is an unchecked exception in Java that is thrown to indicate an illegal or unsuitable argument passed to a method. It is one of the most common exceptions that occur in Java.

Since IllegalArgumentException is an unchecked exception, it does not need to be declared in the throws clause of a method or constructor.

What Causes IllegalArgumentException

An IllegalArgumentException code> occurs when an argument passed to a method doesn’t fit within the logic of the usage of the argument. Some of the most common scenarios for this are:

1. When the arguments passed to a method are out of range. For example, if a method declares an integer age as a parameter, which is expected to be a positive integer. If a negative integer value is passed, an IllegalArgumentException will be thrown.
2. When the format of an argument is invalid. For example, if a method declares a string email as a parameter, which is expected in an email address format.
3. If a null object is passed to a method when it expects a non-empty object as an argument.

IllegalArgumentException Example

Here is an example of a IllegalArgumentException thrown when the argument passed to a method is out of range:

In this example, the main() method calls the setAge() method with the age code> argument set to -1. Since setAge() code> expects age to be a positive number, it throws an IllegalArgumentException code>:

How to Resolve IllegalArgumentException

The following steps should be followed to resolve an IllegalArgumentException in Java:

1. Inspect the exception stack trace and identify the method that passes the illegal argument.
2. Update the code to make sure that the passed argument is valid within the method that uses it.
3. To catch the IllegalArgumentException , try-catch blocks can be used. Certain situations can be handled using a try-catch block such as asking for user input again instead of stopping execution when an illegal argument is encountered.

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java.lang.IllegalArgumentException – How to solve Illegal Argument Exception

Posted by: Sotirios-Efstathios Maneas in exceptions March 20th, 2014 0 Views

In this tutorial, we will discuss how to solve the java.lang.illegalargumentexception – IllegalArgumentException in Java.

This exception is thrown in order to indicate that a method has been passed an illegal or inappropriate argument. For example, if a method requires a non-empty string as a parameter and the input string equals null, the IllegalArgumentException is thrown to indicate that the input parameter cannot be null.

You can also check this tutorial in the following video:

This exception extends the RuntimeException class and thus belongs to those exceptions that can be thrown during the operation of the Java Virtual Machine (JVM). It is an unchecked exception and thus, it does not need to be declared in a method’s or a constructor’s throws clause. Finally, the IllegalArgumentException exists since the first version of Java (1.0).

1. The IllegalArgumentException in Java

The IllegalArgumentException is a good way of handling possible errors in your application’s code. This exception indicates that a method is called with incorrect input arguments. Then, the only thing you must do is correct the values of the input parameters. In order to achieve that, follow the call stack found in the stack trace and check which method produced the invalid argument.

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java.lang.IllegalArgumentException – Reasons and How to Solve?

Here you will learn possible causes of Exception in thread “main” java.lang.IllegalArgumentException and ways to solve it.

I hope you know the difference between error and exception. Error means programming mistake that can be recoverable only by fixing the application code. But exceptions will arise only when exceptional situations occurred like invalid inputs, null values, etc. They can be handled using try catch blocks.

java.lang.IllegalArgumentException will raise when invalid inputs passed to the method. This is most frequent exception in java.

Reasons for java.lang.IllegalArgumentException

Here I am listing out some reasons for raising the illegal argument exception

• When Arguments out of range. For example percentage should lie between 1 to 100. If user entered 200 then illegalarugmentexcpetion will be thrown.
• When arguments format is invalid. For example if our method requires date format like YYYY/MM/DD but if user is passing YYYY-MM-DD. Then our method can’t understand. Then illegalargument exception will raise.
• When closed files has given as argument for a method to read that file. That means argument is invalid.
• When a method needs non empty string as a parameter but null string is passed.

Like this when invalid arguments given then it will raise illegal argument exception.

How to Handle java.lang.IllegalArgumentException?

IllegalArgumentException extends RuntimeException and it is unchecked exception. So we don’t need to catch it. We have to fix the code to avoid this exception.

Hierachy of illegalArgumentException:

• java.lang.Object
  • java.lang.Throwable
    • java.lang.Exception
      • java.lang.RuntimeException
        • java.lang.illegalArgumentException

Example program which will raise illegalArgumentException.

Источник

Illegal Argument Exception Class

Definition

Some information relates to prerelease product that may be substantially modified before it’s released. Microsoft makes no warranties, express or implied, with respect to the information provided here.

Thrown to indicate that a method has been passed an illegal or inappropriate argument.

Portions of this page are modifications based on work created andВ shared by the Android Open Source Project and used according to terms described in the Creative Commons 2.5 Attribution License.

Constructors

Constructs an IllegalArgumentException with no detail message.

A constructor used when creating managed representations of JNI objects; called by the runtime.

Constructs an IllegalArgumentException with the specified detail message.

Constructs a new exception with the specified detail message and cause.

Constructs a new exception with the specified cause and a detail message of (cause==null ? null : cause.toString()) (which typically contains the class and detail message of cause ).

Fields

Properties

Returns the cause of this throwable or null if the cause is nonexistent or unknown.

(Inherited from Throwable) Class (Inherited from Throwable) Handle

The handle to the underlying Android instance.

(Inherited from Throwable) JniIdentityHashCode (Inherited from Throwable) JniPeerMembers LocalizedMessage

Creates a localized description of this throwable.

(Inherited from Throwable) Message

Returns the detail message string of this throwable.

(Inherited from Throwable) PeerReference (Inherited from Throwable) StackTrace (Inherited from Throwable) ThresholdClass

This API supports the Mono for Android infrastructure and is not intended to be used directly from your code.

This API supports the Mono for Android infrastructure and is not intended to be used directly from your code.

Methods

Appends the specified exception to the exceptions that were suppressed in order to deliver this exception.

(Inherited from Throwable) Dispose() (Inherited from Throwable) Dispose(Boolean) (Inherited from Throwable) FillInStackTrace()

Fills in the execution stack trace.

(Inherited from Throwable) GetStackTrace()

Provides programmatic access to the stack trace information printed by #printStackTrace() .

(Inherited from Throwable) GetSuppressed()

Returns an array containing all of the exceptions that were suppressed, typically by the try -with-resources statement, in order to deliver this exception.

(Inherited from Throwable) InitCause(Throwable)

Initializes the cause of this throwable to the specified value.

(Inherited from Throwable) PrintStackTrace()

Prints this throwable and its backtrace to the standard error stream.

(Inherited from Throwable) PrintStackTrace(PrintStream)

Prints this throwable and its backtrace to the specified print stream.

(Inherited from Throwable) PrintStackTrace(PrintWriter)

Prints this throwable and its backtrace to the specified print writer.

(Inherited from Throwable) SetHandle(IntPtr, JniHandleOwnership)

Sets the Handle property.

(Inherited from Throwable) SetStackTrace(StackTraceElement[])

Sets the stack trace elements that will be returned by #getStackTrace() and printed by #printStackTrace() and related methods.

(Inherited from Throwable) ToString() (Inherited from Throwable) UnregisterFromRuntime() (Inherited from Throwable)

Explicit Interface Implementations

Extension Methods

Performs an Android runtime-checked type conversion.

Источник

A quick guide to how to fix IllegalArgumentException in java and java 8?

1. Overview

In this tutorial, We’ll learn when IllegalArgumentException is thrown and how to solve IllegalArgumentException in java 8 programming.

This is a very common exception thrown by the java runtime for any invalid inputs. IllegalArgumentException is part of java.lang package and this is an unchecked exception.

IllegalArgumentException is extensively used in java api development and used by many classes even in java 8 stream api.

First, we will see when we get IllegalArgumentException in java with examples and next will understand how to troubleshoot and solve this problem?

2. Java.lang.IllegalArgumentException Simulation

In the below example, first crated the ArrayList instance and added few string values to it. 

package com.javaprogramto.exception.IllegalArgumentException;

import java.util.ArrayList;
import java.util.List;

public class IllegalArgumentExceptionExample {

	public static void main(String[] args) {
		
		// Example 1
		List<String> list = new ArrayList<>(-10);
		list.add("a");
		list.add("b");
		list.add("c");
	}
}

Output:

Exception in thread "main" java.lang.IllegalArgumentException: Illegal Capacity: -10
	at java.base/java.util.ArrayList.<init>(ArrayList.java:160)
	at com.javaprogramto.exception.IllegalArgumentException.IllegalArgumentExceptionExample.main(IllegalArgumentExceptionExample.java:11)

From the above output, we could see the illegal argument exception while creating the ArrayList instance.

Few other java api including java 8 stream api and custom exceptions.

3. Java IllegalArgumentException Simulation using Java 8 stream api

Java 8 stream api has the skip() method which is used to skip the first n objects of the stream.

public class IllegalArgumentExceptionExample2 {

	public static void main(String[] args) {
		
		// Example 2
		List<String> stringsList = new ArrayList<>();
		stringsList.add("a");
		stringsList.add("b");
		stringsList.add("c");
		
		stringsList.stream().skip(-100);
	}
}

Output:

Exception in thread "main" java.lang.IllegalArgumentException: -100
	at java.base/java.util.stream.ReferencePipeline.skip(ReferencePipeline.java:476)
	at com.javaprogramto.exception.IllegalArgumentException.IllegalArgumentExceptionExample2.main(IllegalArgumentExceptionExample2.java:16)

4. Java IllegalArgumentException — throwing from custom condition

In the below code, we are checking the employee age that should be in between 18 and 65. The remaining age groups are not allowed for any job.

Now, if we get the employee object below 18 or above 65 then we need to reject the employee request.

So, we will use the illegal argument exception to throw the error.

import com.javaprogramto.java8.compare.Employee;

public class IllegalArgumentExceptionExample3 {

	public static void main(String[] args) {

		// Example 3
		Employee employeeRequest = new Employee(222, "Ram", 17);

		if (employeeRequest.getAge() < 18 || employeeRequest.getAge() > 65) {
			throw new IllegalArgumentException("Invalid age for the emp req");
		}

	}
}

Output:

Exception in thread "main" java.lang.IllegalArgumentException: Invalid age for the emp req
	at com.javaprogramto.exception.IllegalArgumentException.IllegalArgumentExceptionExample3.main(IllegalArgumentExceptionExample3.java:13)

5. Solving IllegalArgumentException in Java

After seeing the few examples on IllegalArgumentException, you might have got an understanding on when it is thrown by the API or custom conditions based.

IllegalArgumentException is thrown only if any one or more method arguments are not in its range. That means values are not passed correctly.

If IllegalArgumentException is thrown by the java api methods then to solve, you need to look at the error stack trace for the exact location of the file and line number.

To solve IllegalArgumentException, we need to correct method values passed to it. But, in some cases it is completely valid to throw this exception by the programmers for the specific conditions. In this case, we use it as validations with the proper error message.

Below code is the solution for all java api methods. But for the condition based, you can wrap it inside try/catch block. But this is not recommended.

package com.javaprogramto.exception.IllegalArgumentException;

import java.util.ArrayList;
import java.util.List;

import com.javaprogramto.java8.compare.Employee;

public class IllegalArgumentExceptionExample4 {

	public static void main(String[] args) {

		// Example 1
		List<String> list = new ArrayList<>(10);
		list.add("a");
		list.add("b");
		list.add("c");

		// Example 2
		List<String> stringsList = new ArrayList<>();
		stringsList.add("a");
		stringsList.add("b");
		stringsList.add("c");

		stringsList.stream().skip(2);

		// Example 3
		Employee employeeRequest = new Employee(222, "Ram", 20);

		if (employeeRequest.getAge() < 18 || employeeRequest.getAge() > 65) {
			throw new IllegalArgumentException("Invalid age for the emp req");
		}
		
		System.out.println("No errors");

	}
}

Output:

6. Conclusion

In this article, We’ve seen how to solve IllegalArgumentException in java.

GitHub

IllegalArgumentException

An IllegalArgumentException is thrown in order to indicate that a method has been passed an illegal argument. This exception extends the RuntimeException class and thus, belongs to those exceptions that can be thrown during the operation of the Java Virtual Machine (JVM). It is an unchecked exception and thus, it does not need to be declared in a method’s or a constructor’s throws clause.

Reasons for java.lang.IllegalArgumentException

• When Arguments out of range. For example, the percentage should lie between 1 to 100. If the user entered 101 then an IllegalArugmentExcpetion will be thrown.
• When argument format is invalid. For example, if our method requires date format like YYYY/MM/DD but if the user is passing YYYY-MM-DD. Then our method can’t understand then IllegalArugmentExcpetion will be thrown.
• When a method needs non-empty string as a parameter but the null string is passed.

Example1

public class Student {
   int m;
   public void setMarks(int marks) {
      if(marks < 0 || marks > 100)
         throw new IllegalArgumentException(Integer.toString(marks));
      else
         m = marks;
   }
   public static void main(String[] args) {
      Student s1 = new Student();
      s1.setMarks(45);
      System.out.println(s1.m);
      Student s2 = new Student();
      s2.setMarks(101);
      System.out.println(s2.m);
   }
}

Output

45
Exception in thread "main" java.lang.IllegalArgumentException: 101
at Student.setMarks(Student.java:5)
at Student.main(Student.java:15)

Steps to solve IllegalArgumentException

• When an IllegalArgumentException is thrown, we must check the call stack in Java’s stack trace and locate the method that produced the wrong argument.
• The IllegalArgumentException is very useful and can be used to avoid situations where the application’s code would have to deal with unchecked input data.
• The main use of this IllegalArgumentException is for validating the inputs coming from other users.
• If we want to catch the IllegalArgumentException then we can use try-catch blocks. By doing like this we can handle some situations. Suppose in catch block if we put code to give another chance to the user to input again instead of stopping the execution especially in case of looping.

Example2

import java.util.Scanner;
public class Student {
    public static void main(String[] args) {
        String cont = "y";
        run(cont);
    }
    static void run(String cont) {
        Scanner scan = new Scanner(System.in);
        while( cont.equalsIgnoreCase("y")) {
           try {
               System.out.println("Enter an integer: ");
               int marks = scan.nextInt();
               if (marks < 0 || marks > 100)
                  throw new IllegalArgumentException("value must be non-negative and below 100");
               System.out.println( marks);
            }
            catch(IllegalArgumentException i) {
               System.out.println("out of range encouneterd. Want to continue");
               cont = scan.next();  
               if(cont.equalsIgnoreCase("Y"))
                   run(cont);
               }
          }
     }
}

Output

Enter an integer:
1
1
Enter an integer:
100
100
Enter an integer:
150
out of range encouneterd. Want to continue
y
Enter an integer: