Java lang illegalargumentexception что за ошибка как исправить

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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Last Updated :
02 Feb, 2021

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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