How to use "throws IllegalArgumentException" in method declaration - java

How do you write a method that has "throws IllegalArgumentEception" in the method declaration. Such like this one: If I were to only return d if d>0 otherwise throw an IllegalArgumentException, how would I do that? Do you use try{} and catch{} ?
public double getPrice(double d) throws IllegalArgumentException {
}

You can do that simply at the beginning of the method:
public double getPrice(double d) throws IllegalArgumentException {
if(d <= 0) {
throw new IllegalArgumentException();
}
// rest of code
}
Also the throws IllegalArgumentException is not really needed in the declaration of the method. This must only be done with checked exceptions. But IllegalArgumentException belongs to the unchecked exceptions.
For more information about those I recommend reading this other question.

You should check the condition and if it doesn't meet throw the exception
Sample code:
public double getPrice(double d) throws IllegalArgumentException {
if (d <= 0) {
throw new IllegalArgumentException("Number is negative or 0");
}
//rest of your logic
}
You can learn more about Java Exception here.

Related

Exception is never thrown in the corresponding try block

I have this method:
public int addInt(int x, int y){
try{
if(x<1 || y<1){
throw new InvalidValueExeption();
}
} catch(InvalidValueExeption i){
System.out.println(i);
}
return x+y;
}
InvalidValueExeption is a custom exception. So I wanted to test this:
#Test
public void test(){
AddClass a = new AddClass();
boolean thrown = false;
try{
a.addInt(-2, 3);
} catch(InvalidValueException e){
thrown=true;
}
assertTrue(thrown);
}
I can't run this test, because it says Exception exception.InvalidValueException is never thrown in the corresponding try block.
What am I doing wrong?
Your addInt() method doesn't throw InvalidValueException (*). Inside the method, you do throw it, but you catch it before it can "leave" your method. So, for the outside world, there is no InvalidValueException coming from your method.
Then, correctly the compiler tells you that there's no point in catching the InvalidValueException.
So, instead of immediately catching the exception inside your method, declare the method to throw InvalidValueException:
public int addInt(int x, int y) throws InvalidValueException {
if (x < 1 || y < 1) {
throw new InvalidValueException();
}
return x + y;
}
Rationale:
Exceptions are meant to tell the caller (**) of some method that it couldn't fulfill its task, i.e. your addInt() method is designed to add only positive numbers. And if someone tries it with a number below 1, the method answers with the exception instead of returning a value, thus saying: "Hey, something went wrong, so I can't give you an answer (and the problem description is [the exception with its message and so on])."
( * ) I assume, the missing "c" is just a typo, and you don't have two different exception classes.
( ** ) That's important. I'm not talking about System.out.println(), as that's telling something to the user, not the caller.
If InvalidValueExeption is a checked exception then the compiler will complain because addInt is not declared to throw InvalidValueExeption.
If InvalidValueExeption is not a checked exception then the test will fail because addInt swallows the InvalidValueExeption.
There's also a possible typo in your question: addInt() throws InvalidValueExeption whereas test() tries to catch InvalidValueException. In the former case exception is spelled "Exeption", in the latter case it is spelled "Exception", note the missing "c".
The following approach will work:
public int addInt(int x, int y) {
if (x < 1 || y < 1) {
throw new InvalidValueException();
}
return x + y;
}
#Test(expected = InvalidValueException.class)
public void test(){
AddClass a = new AddClass();
a.addInt(-2, 3);
}
First of i think your InvalidValueExeption is a subtype of RuntimeException.
RuntimeException and its subclasses are unchecked exceptions.
Unchecked exceptions do not need to be declared in a method or
constructor's throws clause if they can be thrown by the execution of
the method or constructor and propagate outside the method or
constructor boundary.
So if you need to indicate that you throw an InvalidValueExeption or inherit Exception instead.
Here the exception is declared on your method and thrown :
public int addInt(int x, int y) throws InvalidValueExeption {
try {
//..
throw new InvalidValueExeption();
} catch (InvalidValueExeption e) {
// do some logging the throw back at you
throw e;
}
}

Handling throws with an ArithmaticException Constructor

Question: Write a method called addTwoPositive that takes two int parameters, and if either value is not positive, throw an ArithmeticException, passing the String "Non-positive integers sent!" to the constructor of the exception. If the values are both positive, then return the sum of them.
I am confused as to how to handling this exception through the ArithmeticException constructor. Here is my code:
package IfnotPos;
public class IfNotPositive extends ArithmeticException{
public static void main(String[] args) throws ArithmeticException{
IfNotPositive pos = new IfNotPositive();
System.out.println(pos.addTwoPositive(-3,2));
}
public int addTwoPositive(int plus, int plus1) throws ArithmeticException{
if(plus < 0 || plus1 < 0){
throw new ArithmeticException("Non-positive integers sent!");
}
else{
return plus + plus1;
}
}
public ArithmeticException(String string) {
return string;
}
}
I get an error of "return type for the method is missing" and if I change it to string or anything else, it will obviously stop being a constructor. Any help handling this exception will be greatly appreciated.
First of all you don't need to extend ArithmeticException class. You are not creating a custom exception, but writing a program to throw ArithmeticException (note this resides in java.lang) if a condition is not met.
Problem with your code is that constructor you have defined is never meant to be there and is not correct, hence the compiling issue.
public ArithmeticException(String string) {
return string;
}
Just remove this constructor and you are good to go.
I don't understand why you want to create constructor for ArithmeticException in your code?
public ArithmeticException(String string) {
return string;
}
You don't write a constructor like this. This is syntax exception complaining about the above method. This constructor is unnecessary. Remove this your code will work perfect. You should see below exception running your program...
Exception in thread "main" java.lang.ArithmeticException: Non-positive
integers sent! at
com.code.samples.IfNotPositive.addTwoPositive(IfNotPositive.java:14)
at com.code.samples.IfNotPositive.main(IfNotPositive.java:8)
From Providing Constructors for Your Classes
Constructor declarations look like method declarations—except that they use the name of the class and have no return type.
I get an error of "return type for the method is missing"
You are getting this error because public ArithmeticException(String string) line in your code.
You can't add constructor of any other class inside another class, so Java treating public ArithmeticException(String string) as function rather constructor.
And as you know for method you need to have return type mendetry
You don't have to extend, ArithmeticException class. See below code:
public class IfNotPositive{
public static void main(String[] args) throws ArithmeticException{
IfNotPositive pos = new IfNotPositive();
System.out.println(pos.addTwoPositive(-3,2));
}
public int addTwoPositive(int plus, int plus1) throws ArithmeticException{
if(plus < 0 || plus1 < 0){
throw new ArithmeticException("Non-positive integers sent!");
}
else{
return plus + plus1;
}
}
}
Q: throw an ArithmeticException, passing the String "Non-positive integers sent!"
It means throw ArithmeticException with detail message "Non-positive integers sent!"
So for that you can use public ArithmeticException(String s) Constructor.
Constructs an ArithmeticException with the specified detail message.
Ex:
throw new ArithmeticException("Non-positive integers sent!");

How to assert that an exception is caught?

I know how to assert that an Exception is thrown. But, how can I assert that an Exception was thrown and it was successfully caught? For example, say I have a method which should throw a particular type of exception for invalid inputs
public static void myMethod(String value) {
try {
someExternalMethod(value);// can throw IllegalArgumentException
} catch (IllegalArgumentException e) {
System.out.println("Let me handle it differently");
} catch (Exception e) {
System.out.println("Not IllegalArgumentException");
}
}
Now I want to assert that for some values the method indeed has thrown 'IllegalArgumentException' and not some other Exception.
In the context of testing myMethod you cannot (and more importantly, you should not want to) check that someExternalMethod has thrown IllegalArgumentException. In fact, your tests of myMethod should not assume that a call of someExternalMethod has been made: it is an implementation detail of myMethod.
The very reason myMethod catches these exceptions is to hide them from its callers. You should check that these exceptions are hidden from you by passing values that cause them, and verifying that nothing is thrown in both cases.
Testing someExternalMethod, along with the exceptions that it throws, is the task accomplished by testing someExternalMethod, not myMethod.
You're missing the important point of unit testing - tests should test behaviour, not implementation.
Given this assumption you should be testing the behaviour of myMethod is as expected when an IllegalArgumentException occurs. It's hard to say any more than that with the method you've shown given the parameter, a single String is immutable, there is no return value and no exceptions are thrown.
A better example might be this method (which is a little contrived to demonstrate the point):
public double divide(int numerator, int denominator)
{
try
{
return numerator / denominator;
}
catch (ArithmeticException e)
{
return Double.NaN;
}
}
Where your tests would assert that the division is correct and that when an error occurs NaN is returned, like this:
#Test
public void testDivide()
{
assertEquals(2.0, divide(4, 2), 0);
}
#Test
public void testDivideByZero()
{
assertTrue(Double.isNaN(divide(1, 0));
}
You could then re-write the divide method like this:
public double divide(int numerator, int denominator)
{
if (denominator == 0)
{
return Double.NaN;
}
else
{
return numerator / denominator;
}
}
And the tests would confirm the operation of my system because the behaviour of the divide method remains unchanged.
catch blocks are evaluated in the order they are.
Your code works fine: in case of IllegalArgumentException, Exception block will be ignored.
Mock method someExternalMethod(value) to force throw Exception
Test method myMethod, checking that is not throwing Exception:
#Test
public void testMyMethod() {
try {
myMethod("value");
} catch (Exception ex) {
Assert.fail();
}
}

How can I throw a general exception in Java?

Consider this simple program. The program has two files:
File Vehicle.java
class Vehicle {
private int speed = 0;
private int maxSpeed = 100;
public int getSpeed()
{
return speed;
}
public int getMaxSpeed()
{
return maxSpeed;
}
public void speedUp(int increment)
{
if(speed + increment > maxSpeed){
// Throw exception
}else{
speed += increment;
}
}
public void speedDown(int decrement)
{
if(speed - decrement < 0){
// Throw exception
}else{
speed -= decrement;
}
}
}
File HelloWorld.java
public class HelloWorld {
/**
* #param args
*/
public static void main(String[] args) {
Vehicle v1 = new Vehicle();
Vehicle v2 = new Vehicle();
// Do something
// Print something useful, TODO
System.out.println(v1.getSpeed());
}
}
As you can see in the first class, I have added a comment ("// throw exception") where I would like to throw an exception. Do I have to define my own class for exceptions or is there some general exception class in Java I can use?
You could create your own Exception class:
public class InvalidSpeedException extends Exception {
public InvalidSpeedException(String message){
super(message);
}
}
In your code:
throw new InvalidSpeedException("TOO HIGH");
You could use IllegalArgumentException:
public void speedDown(int decrement)
{
if(speed - decrement < 0){
throw new IllegalArgumentException("Final speed can not be less than zero");
}else{
speed -= decrement;
}
}
Well, there are lots of exceptions to throw, but here is how you throw an exception:
throw new IllegalArgumentException("INVALID");
Also, yes, you can create your own custom exceptions.
A note about exceptions. When you throw an exception (like above) and you catch the exception: the String that you supply in the exception can be accessed throw the getMessage() method.
try{
methodThatThrowsException();
}catch(IllegalArgumentException e)
{
e.getMessage();
}
It really depends on what you want to do with that exception after you catch it. If you need to differentiate your exception then you have to create your custom Exception. Otherwise you could just throw new Exception("message goes here");
The simplest way to do it would be something like:
throw new java.lang.Exception();
However, the following lines would be unreachable in your code. So, we have two ways:
Throw a generic exception at the bottom of the method.
Throw a custom exception in case you don't want to do 1.
Java has a large number of built-in exceptions for different scenarios.
In this case, you should throw an IllegalArgumentException, since the problem is that the caller passed a bad parameter.
You can define your own exception class extending java.lang.Exception (that's for a checked exception - these which must be caught), or extending java.lang.RuntimeException - these exceptions does not have to be caught.
The other solution is to review the Java API and finding an appropriate exception describing your situation: in this particular case I think that the best one would be IllegalArgumentException.
It depends. You can throw a more general exception, or a more specific exception. For simpler methods, more general exceptions are enough. If the method is complex, then, throwing a more specific exception will be reliable.

exceptions in java

I wrote a code which checks all kinds of conditions.
If it meets the condition it does what it is supposed to, otherwise I want it to throw an
exception.
Is there any special syntax for that? Otherwise the compiler wants me to return any array,
which I don't want to, due to the pre-condition.
Here is part of my code:
public static int [] code(int[]arr){
if ((arr!=null)&&(chack4and5(arr))&&(arr[arr.length-1]!=4)&&(TwoFours(arr))){
int k=0;
for(int i = 0; i<=arr.length-1; i++){
if (arr[i] == 4){
int place= pos(arr,k);
arr[place]=arr[i+1];
arr[i+1]=5;
k=k+3;
}
}
return arr;
}
else {
System.out.println("Please enter a legal array which matches the pre- conditions");
}
}
}
The way to throw an exception is
throw new IllegalArgumentException(
"Please enter a legal array which matches the pre- conditions");
IllegalArgumentException is a Java runtime exception suitable for the current situation, but of course you can choose another one, or create and use your own type too. The only restriction is that it must be a subclass of java.lang.Exception.
I would rearrrange your code though to check the preconditions first, then proceed if everything's fine - I find this more readable:
if (arr == null || !chack4and5(arr) || arr[arr.length-1] == 4 || !TwoFours(arr)) {
throw new IllegalArgumentException(
"Please enter a legal array which matches the pre- conditions");
}
int k=0;
for(int i = 0; i<=arr.length-1; i++){
if (arr[i] == 4){
int place= pos(arr,k);
arr[place]=arr[i+1];
arr[i+1]=5;
k=k+3;
}
}
return arr;
(In fact, I would even prefer extracting the precondition check into a separate method - but I leave this to you.)
throw new IllegalArgumentException(
"Please enter a legal array which matches the pre- conditions")
java.langIllegalArgumentException is a RuntimeException that means some of the arguments are not as they are expected to be. Since it is an unchecked exceptions, your callers are not forced to handle it in any way (as opposed to checked exceptions)
You can throw an Exception by yourself. Maybe the best way to do this is defining a custom exception and then throwing it. If you don't want to do that use an IllegalArgumentException.
Here an example of a custom exception:
public static int [] code(int[]arr) {
if ((arr!=null)&&(chack4and5(arr))&&(arr[arr.length-1]!=4)&&(TwoFours(arr))){
int k=0;
for(int i = 0; i<=arr.length-1; i++){
if (arr[i] == 4){
int place= pos(arr,k);
arr[place]=arr[i+1];
arr[i+1]=5;
k=k+3;
}
}
return arr;
}
else {
throw new MyException("No legal array");
}
}
}
And here your custom exception:
public class MyException extends Exception {
public MyException(String message) {
super(message);
}
}
If the exception is that something about your arguments is illegal, then throw an IllegalArgumentException:
throw new IllegalArgumentException("descriptive message")
You may want to take a look at Oracle's tutorials on exceptions.
To throw an exception, you use the throw keyword.
To mark that a method may throw an exception, use the throws keyword, like
public static void foo() throws SomeException
You can throw exception with this line
throw new SomeKindOfException("Exception description"); // or any other exception, also yours...
But you need to specify at the method declaration:
public static int [] code(int[]arr) throws SomeKindOfException{
See Oracle tutorial for more

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