problem with try catch

[naved momin]

what i want is when the marks are above 100 then catch clause should be executed
code is

import java.io.*;
class CalPer
{
	int semarks;
	int osmarks;

	void setper()throws Exception
	{
		BufferedReader b = new BufferedReader(new InputStreamReader(System.in));
		
		
		System.out.print(" enter your se marks : ");
		semarks = Integer.parseInt(b.readLine());
		System.out.print(" enter your os marks : ");
		osmarks = Integer.parseInt(b.readLine());
		
		System.out.println(" your os marks are : "  + osmarks + " " + " your se marks are : "  + semarks  );
		
		int result = 0;
		try
		{
			if( (semarks <= 100) && (osmarks <= 100 ) )
			{
				result = semarks + osmarks / 200 * 100 ;	
		
				System.out.println(" your calculated percentage is :  " + result + " "+ "%");
			}
	
		}
		
		catch(NumberFormatException e)
		{
			System.err.println(" please enter valid no between 0 to 100 ");
		}
	}
	
	public static void main(String[] args)throws Exception
	{
		CalPer cp = new CalPer();
		cp.setper();
	}
}

output is

[naseem@naseem exception]$ java CalPer
 enter your se marks : 102
 enter your os marks : 105
 your os marks are : 105  your se marks are : 102

[Joanne Neal]

naved momin wrote:what i want is when the marks are above 100 then catch clause should be executed

Then add an else statement that throws a NumberFormatException, but it would be better to just add an else statement that printed the error message.

[Campbell Ritchie]

You will only get a NumberFormatException from that code if you enter something like 1.23 or Campbell. If what you enter is a valid integer, you will not get an Exception.
What you want is an IllegalArgumentException.

But you ought not to throw an Exception and catch it in the same method. Rather than doing that, you want an if-else, preferably wrapped inside a loop.
And why are you not using a Scanner, nearly 7 years after it was introduced?

[Piyush Joshi]

naved momin wrote:what i want is when the marks are above 100 then catch clause should be executed

This is not an appropriate use of NumberFormatException.
NumberFormatException should be "Thrown to indicate that the application has attempted to convert
a string to one of the numeric types, but that the string does not have the appropriate format."

Instead of this your logic can be implemented by writing an else block to your if block. For Example:

try {
			System.out.print(" enter your se marks : ");
			semarks = Integer.parseInt(b.readLine()); // NumberFormatException can be thrown from here
			System.out.print(" enter your os marks : ");
			osmarks = Integer.parseInt(b.readLine()); // NumberFormatException can be thrown from here

			System.out.println(" your os marks are : " + osmarks + " "
					+ " your se marks are : " + semarks);

			int result = 0;

			if ((semarks <= 100) && (osmarks <= 100)) {
				result = semarks + osmarks / 200 * 100;

				System.out.println(" your calculated percentage is :  "
						+ result + " " + "%");
			}
			else{
				System.err.println(" please enter valid no between 0 to 100 ");
			}

		}

		catch (NumberFormatException e) {
			System.err.println(" please enter valid no between 0 to 100 ");
		}

Oh!! lost to Campbell

[Campbell Ritchie]

No, lost to Joanne.

[Campbell Ritchie]

. . . and what are you doing if a negative mark is entered?

[naved momin]

Piyush Joshi wrote:

naved momin wrote:what i want is when the marks are above 100 then catch clause should be executed

This is not an appropriate use of NumberFormatException.
NumberFormatException should be "Thrown to indicate that the application has attempted to convert
a string to one of the numeric types, but that the string does not have the appropriate format."

Instead of this your logic can be implemented by writing an else block to your if block. For Example:

try {
			System.out.print(" enter your se marks : ");
			semarks = Integer.parseInt(b.readLine()); // NumberFormatException can be thrown from here
			System.out.print(" enter your os marks : ");
			osmarks = Integer.parseInt(b.readLine()); // NumberFormatException can be thrown from here

			System.out.println(" your os marks are : " + osmarks + " "
					+ " your se marks are : " + semarks);

			int result = 0;

			if ((semarks <= 100) && (osmarks <= 100)) {
				result = semarks + osmarks / 200 * 100;

				System.out.println(" your calculated percentage is :  "
						+ result + " " + "%");
			}
			else{
				System.err.println(" please enter valid no between 0 to 100 ");
			}

		}

		catch (NumberFormatException e) {
			System.err.println(" please enter valid no between 0 to 100 ");
		}

Oh!! lost to Campbell

my refined code is

import java.io.*;
class CalPer
{
	int semarks;
	int osmarks;

	void setper()throws Exception
	{
		BufferedReader b = new BufferedReader(new InputStreamReader(System.in));
		
		
		System.out.print(" enter your se marks : ");
		semarks = Integer.parseInt(b.readLine());
		System.out.print(" enter your os marks : ");
		osmarks = Integer.parseInt(b.readLine());
		
		System.out.println(" your os marks are : "  + osmarks + " " + " your se marks are : "  + semarks  );
		
		int result = 0;
		try
		{
			if( (semarks <= 100) && (semarks >=0) && (osmarks <= 100 ) && (osmarks >=0 ))
			{
				result = semarks + osmarks / 200 * 100 ;	
		
				System.out.println(" your calculated percentage is :  " + result + " "+ "%");
			
			}
			else
			{
				System.err.println("[" + osmarks +"] [ " + semarks + "] you are not supposed to enter negative marks ");
			}
	
		}
		
		catch(NumberFormatException e)
		{
			System.err.println(" this is not a integer value please enter integer value only . ");
		}
	}
	
	public static void main(String[] args)throws Exception
	{
		CalPer cp = new CalPer();
		cp.setper();
	}
}

sorry what i want to do is when some one entered float value or when some one entered something like fortyfive for marks then this should be catch by catch clause ....i m learning try catch module from my book , so i m not a experienced programmer or so
can you tell me what should be done for that ?

[Piyush Joshi]

Campbell Ritchie wrote:No, lost to Joanne.

Did'nt see that one also!!

Campbell Ritchie wrote:. . . and what are you doing if a negative mark is entered?

Its upto naved.

[naved momin]

Piyush Joshi wrote:

Campbell Ritchie wrote:No, lost to Joanne.

Did'nt see that one also!!

Campbell Ritchie wrote:. . . and what are you doing if a negative mark is entered?

Its upto naved.

but now whenever i entered a string like fortyfive which could be possible input so this should be catch by catch clause how to do that , now it is giving me a runtime exception of numberformateception ?
runtime exception is

Exception in thread "main" java.lang.NumberFormatException: For input string: "thee"
	at java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
	at java.lang.Integer.parseInt(Integer.java:449)
	at java.lang.Integer.parseInt(Integer.java:499)
	at CalPer.setper(CalPer.java:13)
	at CalPer.main(CalPer.java:45)

[Piyush Joshi]

naved momin wrote:
sorry what i want to do is when some one entered float value or when some one entered something like fortyfive for marks then this should be catch by catch clause ....i m learning try catch module from my book , so i m not a experienced programmer or so
can you tell me what should be done for that ?

Do you understand why NumberFormatException or for that matter some exception is thrown?
Its because of some error for example: invalid arguments passed to methods.

Next you need to understand What does the following code do:

semarks = Integer.parseInt(b.readLine());

b.readLine() reads whatever user entered and returns it as a String.
and then this String is passed to parseInt() method in Integer class. If you read the javadocs for this method
then you can see that this method throws NumberFormatException if if the string does not contain a parsable integer.

Therefore if some one enters float value or something like fortyfive and when you parse it to Integer then a NumberFormatException will be thrown.

  System.out.print(" enter your se marks : ");
  semarks = Integer.parseInt(b.readLine()); // NumberFormatException will be thrown from here  if invalid value is entered
  // here invalid means values which can not be parsed into Integer
  System.out.print(" enter your os marks : ");
  osmarks = Integer.parseInt(b.readLine()); // NumberFormatException will be thrown from here  if invalid value is entered

Now If you want to handle this exception i.e. you want to do something like print an error msg then you should enclose the above method calls in a try block and write a following catch block catching NumberFormatException, like this:

     try {
			 System.out.print(" enter your se marks : ");
                         semarks = Integer.parseInt(b.readLine()); // NumberFormatException will be thrown from here  if invalid value is entered
                         // here invalid means values which can not be parsed into Integer
                         System.out.print(" enter your os marks : ");
                         osmarks = Integer.parseInt(b.readLine()); // NumberFormatException will be thrown from here  if invalid value is entered
                        
                       //If an exception is thrown then following code will not execute
			System.out.println(" your os marks are : " + osmarks + " "
					+ " your se marks are : " + semarks);

			int result = 0;

			if ((semarks <= 100) && (osmarks <= 100)) {
				result = semarks + osmarks / 200 * 100;

				System.out.println(" your calculated percentage is :  "
						+ result + " " + "%");
			}
			else{
				System.err.println(" please enter valid no between 0 to 100 ");
			}

		}
		catch (NumberFormatException e) {
                //If an NumberFormatException is thrown then following code will execute
			System.err.println(" please enter valid no between 0 to 100 ");
		}
	}

I hope it answers your doubts.

[naved momin]

Piyush Joshi wrote:

naved momin wrote:
sorry what i want to do is when some one entered float value or when some one entered something like fortyfive for marks then this should be catch by catch clause ....i m learning try catch module from my book , so i m not a experienced programmer or so
can you tell me what should be done for that ?

Do you understand why NumberFormatException or for that matter some exception is thrown?
Its because of some error for example: invalid arguments passed to methods.

Next you need to understand What does the following code do:

semarks = Integer.parseInt(b.readLine());

b.readLine() reads whatever user entered and returns it as a String.
and then this String is passed to parseInt() method in Integer class. If you read the javadocs for this method
then you can see that this method throws NumberFormatException if if the string does not contain a parsable integer.

Therefore if some one enters float value or something like fortyfive and when you parse it to Integer then a NumberFormatException will be thrown.

  System.out.print(" enter your se marks : ");
  semarks = Integer.parseInt(b.readLine()); // NumberFormatException will be thrown from here  if invalid value is entered
  // here invalid means values which can not be parsed into Integer
  System.out.print(" enter your os marks : ");
  osmarks = Integer.parseInt(b.readLine()); // NumberFormatException will be thrown from here  if invalid value is entered

Now If you want to handle this exception i.e. you want to do something like print an error msg then you should enclose the above method calls in a try block and write a following catch block catching NumberFormatException, like this:

     try {
			 System.out.print(" enter your se marks : ");
                         semarks = Integer.parseInt(b.readLine()); // NumberFormatException will be thrown from here  if invalid value is entered
                         // here invalid means values which can not be parsed into Integer
                         System.out.print(" enter your os marks : ");
                         osmarks = Integer.parseInt(b.readLine()); // NumberFormatException will be thrown from here  if invalid value is entered
                        
                       //If an exception is thrown then following code will not execute
			System.out.println(" your os marks are : " + osmarks + " "
					+ " your se marks are : " + semarks);

			int result = 0;

			if ((semarks <= 100) && (osmarks <= 100)) {
				result = semarks + osmarks / 200 * 100;

				System.out.println(" your calculated percentage is :  "
						+ result + " " + "%");
			}
			else{
				System.err.println(" please enter valid no between 0 to 100 ");
			}

		}
		catch (NumberFormatException e) {
                //If an NumberFormatException is thrown then following code will execute
			System.err.println(" please enter valid no between 0 to 100 ");
		}
	}

I hope it answers your doubts.

a million thanks !