Conversion of string into integer in Java

Code Java:

---

public static void main(String[] args) {
 
String yourString = "420";
int yourInt = Integer.parseInt(yourString);
 
System.out.println(yourInt);
}

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Dude you just help me big time now i can finish like 7 programs!:-bd:-bd:-bd:-bd

If you are unsure of the input, make sure to catch any NumberFormatExceptions that can be thrown when trying to parse something that can't be parsed

Code Java:

---

public static void main(String[] args) {
     int yourInt = -1;
    if ( args.length > 0 ){     
        try{
            yourInt = Integer.parseInt(args[0]);
        }catch ( NumberFormatException e ){
            System.out.println("Wrong number...");
            return;
        }
    }else{
        System.out.println("No input....");
        return;
    }
    System.out.println(yourInt);
}

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Personally I'd check the input before trying to parse it as exceptions are slow.

// Json

Quote:

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Originally Posted by Json

Personally I'd check the input before trying to parse it as exceptions are slow.

// Json

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Good point. Out of curiosity though...according to this benchmark below going through the try/catch block is 10 times faster than validation (with correct input of course). For incorrect input the validation is 2 times faster than try/catch. I know in the real world things are much different, and there are probably other ways that are potentially faster to do validation, but its just food for thought.

Code Java:

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public class ExceptionTest{
 
	public static void main( String[] args ){
		String input = "126";
		long start = System.currentTimeMillis();
 
		for ( int i = 0 ; i < 100000; i++ ){
			try{
				int st = Integer.parseInt(input);
			}catch ( NumberFormatException e ){
				//System.out.println("Error");
			}
		}
 
		System.out.println(System.currentTimeMillis() - start );
		start = System.currentTimeMillis();
		for ( int i = 0 ; i < 100000; i++ ){
			if ( !input.matches("[0-9]+") ){
				//System.out.println("Error");
			}else{
                               int st = Integer.parseInt(input);
                        }
 
		}
		System.out.println(System.currentTimeMillis() - start );
	}
}

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I updated your code somewhat to add an input checker of my own.

Code Java:

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package uk.co.cdl.testing.io;
 
public class ExceptionTest1 {
    public static void main(String[] args) {
        String input = "126h";
        long start = System.currentTimeMillis();
 
        for (int i = 0; i < 100000; i++) {
            try {
                int st = Integer.parseInt(input);
            } catch (NumberFormatException e) {
                //System.out.println("Error");
            }
        }
 
        System.out.println("Exceptions: " + (System.currentTimeMillis() - start) + "ms");
        start = System.currentTimeMillis();
        for (int i = 0; i < 100000; i++) {
            if (!input.matches("[0-9]+")) {
                //System.out.println("Error");
            } else {
                int st = Integer.parseInt(input);
            }
 
        }
        System.out.println("Regexp: " + (System.currentTimeMillis() - start) + "ms");
 
 
        start = System.currentTimeMillis();
        for (int i = 0; i < 100000; i++) {
            if (isDigits(input)) {
                int st = Integer.parseInt(input);
            }
        }
        System.out.println("isDigits: " + (System.currentTimeMillis() - start) + "ms");
    }
 
    public static boolean isDigits(final String input) {
        if (input == null) {
            return false;
        } else if ("".equals(input)) {
            return false;
        } else {
            for (int i = 0; i < input.length(); ++i) {
                if (!Character.isDigit(input.charAt(i))) {
                    return false;
                }
            }
        }
 
        return true;
    }
}

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In the case where the input is not a number I get the following output.

Quote:

---

Exceptions: 203ms
Regexp: 141ms
isDigits: 0ms

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And in the case were the input is a number I get the following output.

Quote:

---

Exceptions: 0ms
Regexp: 141ms
isDigits: 0ms

---

// Json

Thanks for posting that code Json.

when i tried this code

Code Java:

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String yourString = "Ashar";
int yourInt = Integer.parseInt(yourString);
 
System.out.println(yourInt);

---

it compiled showing no errors ... however when I run the program i get an exception in thread "main" java.lang.NumberFormatException: for input String "Ashar" along with other things on the console after running the program

I want to know why that happens ... I thought this statement will give me the ASCII equivalent to the string "Ashar"

Quote:

---

Originally Posted by Ashar

when i tried this code

Code Java:

---

String yourString = "Ashar";
int yourInt = Integer.parseInt(yourString);
 
System.out.println(yourInt);

---

it compiled showing no errors ... however when I run the program i get an exception in thread "main" java.lang.NumberFormatException: for input String "Ashar" along with other things on the console after running the program

I want to know why that happens ... I thought this statement will give me the ASCII equivalent to the string "Ashar"

---

No this is designed for converting a String representation of a Numerical value into a numerical value.

I.e "1" would become integer 1 and "8" would become integer 8 rather than string literals.

the following would print each of the ascii values.

Code Java:

---

public void printAscii(String s){
   for(int i = s.length-1; i > -1; i--)
      System.out.print((int)s.charAt(i));
   System.out.println();
}

---

Chris

Thank you chris :)

can you tell me why was it used in a program to read a number .. does JOptioPane always reads strings?
like this :

Code Java:

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 int x = Integer parseInt(JOprionPane.showInputDialog("Enter number") ;

---

JOptionPane.showInputDialog() does indeed always return a String

Or null?

// Json

Well yes or null i suppose, but its a null String lol ;)

hehe :D

// Json

Hi guys, I don't understand a thing...
Let me give you some code first:

Code Java:

---

public class Airplane {
 
	private String plane, firstLetter, lastLetter;
	private int nRows;
	private int columns;
	String[] windowsSeats;
	String[] aisleSeats;
 
 
 
	public Airplane(String plane, int nRows, String firstLetter, String lastLetter) {
		this.plane=plane;
		this.nRows=nRows;
		this.firstLetter=firstLetter;
		this.lastLetter=lastLetter;
 
 
		try{
			this.columns=Integer.parseInt(lastLetter)-Integer.parseInt(firstLetter)+1;
        }catch ( NumberFormatException e ){
            System.out.println("Wrong number...");
            return;
        }
...
 
	}

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I throw it with a kind of main like:

Code Java:

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public class MainExample {
 
 
	public static void main(String[] args) throws FullFlight, SeatTaken, NoPlane, NoSeat {
 
 
		FlightManager  fm = new FlightManager();
		fm.addAirplane("B737-1", 32, "A", "F");
		fm.addAirplane("B737-2", 42, "A", "F");
 
		...
}

---

I obtain always the error message.. Can you explain why the conversion (this.columns..) doesn't succeed?

Hello there, its because you are trying to find a numeric value in a string which is the letter "F" or "A", this will always throw a NumberFormatException because neither A or F are numbers.

If you wish to convert this into their ASCII equivalents you can use this.

Code Java:

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Character.getNumericValue(lastLetter);

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You also need to make sure that lastLetter and firstLetter is a char instead of a String.

Hope that helps.

// Json

Hi Json and thanks for the answer first!
I understood your thesis but unfortunately I can not change the type of firstLetter and lastLetter. they're passed as String in the function.

Can I find another way to solve it?
Take into account that it is a Flight Manager program, the function "addAirplane" receives these 2 String as parameters, and I must be able to, when I make a reservation, understand how many "columns" of seats there are and also for compute the capacity of an airplane (rows*columns).

I accept any kind of explanation and help.
Is very important to me to understand because it is part of an exam at univerisity.

Thanks

If you know that the 2 strings are always going to be 1 character long you can do this.

Code Java:

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char character = firstLetter.charAt(0);

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That will grab the character at the first position in the string and then you can use that to get the int value.

// Json