Two Dimensional array declaration in Java

[Tanka Banerjee]

Hi,
I have wrritten a two dimensional array in Java.
Let's check the program.
It is not giving any compilation error but there is runtime error NumberFormatException.

Regards
Tanka

import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
class multidimensionalarray {
int arr[][];
 
void inarr(int b)
{
try{
for(int i=0;i<4;i++)
 {
 for(int j=0;j<4;j++)
arr[i][j] =b;
}
}
catch(Exception e)
{
e.printStackTrace();
}
 
}
// end of inarr()
 
public static void main(String args[]) throws IOException {
multidimensionalarray mdar = new multidimensionalarray();
BufferedReader br1 = new BufferedReader(new InputStreamReader(System.in));
String br2=br1.readLine();
int br=Integer.parseInt(br2);
mdar.inarr(br);
//System.out.println(arr[i][j]);
}
// end of main()
}

[Campbell Ritchie]

Why are you putting a try-catch around the code in inarr which doesn't declare an exception, and not catching the IOException from the BufferedReader? And why are you not closing the BufferedReader?

What are you entering to get the NumberFormatException? You must enter something which is a real number, then you won't get the NumberFormatException, so you can go on and deal with the NullPointerException which you will get in the next stage

[Campbell Ritchie]

By the way: there is no such thing as a two dimensional array in Java. What you have set up is an array of arrays.

[Tanka Banerjee]

Hi,
try-catch is not necessary in the code.
IOException is being thrown.
NumberFormatException comes when I give input like:
java multidimensionalarray 4 5 6 7 7 9 7 6
6 8 6 78 9 7
at java.lang.NumberFormatException.forInputString(NumberFormatException.
java:48)
at java.lang.Integer.parseInt(Integer.java:456)
at java.lang.Integer.parseInt(Integer.java:497)
at multidimensionalarray.main(multidimensionalarray.java:28)

Regards
Tanka


[ December 23, 2008: Message edited by: Tanka Banerjee ]

[Tanka Banerjee]

Originally posted by Campbell Ritchie:
Why are you putting a try-catch around the code in inarr which doesn't declare an exception, and not catching the IOException from the BufferedReader? And why are you not closing the BufferedReader?

What are you entering to get the NumberFormatException? You must enter something which is a real number, then you won't get the NumberFormatException, so you can go on and deal with the NullPointerException which you will get in the next stage

I need to change the logic.

[Tanka Banerjee]

java multidimensionalarray 4.0 6.9
4.5 76.9
Exception in thread "main" java.lang.NumberFormatException: For input string: "4
.5 76.9 "
at java.lang.NumberFormatException.forInputString(NumberFormatException.
java:48)
at java.lang.Integer.parseInt(Integer.java:456)
at java.lang.Integer.parseInt(Integer.java:497)
at multidimensionalarray.main(multidimensionalarray.java:28)
I am getting the NumberFormatException when I am typing this input.

Regards
Tanka.

[Campbell Ritchie]

You are trying to convert 4.5 to an int. 4.5 isn't an int.

[Campbell Ritchie]

And if you are using a BufferedReader, you can only enter one number on each line. A Scanner would have been easier to use.

Why are you passing command-line arguments? You don't use them anywhere in the code. It seems a very strange bit of design to use "try" round code which doesn't declare an Exception and "throws" on code which does declare an Exception.

[Tanka Banerjee]

Code Modified.

import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
class multidimensionalarray {
int[][] arr= new int[4][4];
 
void inarr(int b)
{
 
for(int i=0;i<4;i++)
 {
 for(int j=0;j<4;j++)
arr[i][j] =b;
}
 
/*catch(Exception e)
{
e.printStackTrace();
}*/
 
}
// end of inarr()
 
public static void main(String args[]) throws IOException {
multidimensionalarray mdar = new multidimensionalarray();
BufferedReader br1 = new BufferedReader(new InputStreamReader(System.in));
String br2=br1.readLine();
int br=Integer.parseInt(br2);
mdar.inarr(br);
br1.close();
//System.out.println(arr[i][j]);
}
// end of main()
}

[ December 23, 2008: Message edited by: Tanka Banerjee ]

[Tanka Banerjee]

java multidimensionalarray 4
4

No error coming.
I will take System.out.println() to see what is happening inside.

Regards
Tanka

[Campbell Ritchie]

You have obviously worked out why you might get a NullPointerException and corrected that.

[ramya narayanan]

Dear All,
The first time I saw the program I didn't notice anything.
But when compiling & running I noticed NullpointerException it was when I realized that the array has not been initialized properly

  
C:\sai>javac Multidimensionalarray.java
 
C:\sai>java Multidimensionalarray
12
[B]java.lang.NullPointerException[/B]
        at Multidimensionalarray.inarr(Multidimensionalarray.java:13)
        at Multidimensionalarray.main(Multidimensionalarray.java:29)
 
  

I rectified the same & found the application running.

But when Campbell ritchie sees that with his vast experience, he was able to find some flaws in the code & offered unique solutions.
The things which I missed out were

1) Unnecessary wrapping up of code in try , catch block when that block of code doesn't have the potential to cause any exception.

2) For any streams or connections, it has to be closed which is not the case here .

But regarding throwing any exceptions from BufferedReader, it has to be kept in mind that this exception is declared in the throws clause.

By the way: there is no such thing as a two dimensional array in Java. What you have set up is an array of arrays.

Fitting reply.

And if you are using a BufferedReader, you can only enter one number on each line. A Scanner would have been easier to use.

How you are saying that?
You mean we can pass the values as commandline arguments & then use Scanner.useDelimiter property to parse them right.
But Scanner's constructor need a file as parameter right.
Then how can we get values & pass it to scanner.

But finally, it was a very good learning experience to learn about the best practices from experienced people like Campbell Ritchie.(Eventhough he is harsh in his reply, atleast it sounds like that)
Thanks Ritchie.
Regards.

[Campbell Ritchie]

I am very sorry for sounding harsh.

You can pass the numbers as command-line arguments, then they are a String array, and you can parse its members. You have seen that array hundreds of times and never noticed it. It is called args.

java DoubleArrayFiller 3

public class DoubleArrayFiller
{
  public static void main(String[] args)
  {
    . . . new DoubleArrayFiller(Integer.parseInt(args[0]));
    . . .

If you forget the command-line argument you will get an ArrayIndexOutOfBoundsException with the number 0. If you pass 3.0 you will get a NumberFormatException.

You can write 2 3 4 5 6 (not at the command line) and pass that line to a Scanner, and use its hasNextInt() and nextInt() methods in a while loop to run the whole app five times, using 2 3 4 5 and 6. Scanner won't work in Java 1.4.2 or earlier.

[ramya narayanan]

You can write 2 3 4 5 6 (not at the command line) and pass that line to a Scanner

Are you saying i need to write a text file with 2 3 4 5 6 & pass that to scanner which takes these values.
Regards.

[Campbell Ritchie]

You can write a text file, or you can pass System.in to the Scanner constructor. Whichever you prefer.

[ramya narayanan]

Using System.in Scanner i'm able to get the int values.

 
import java.util.Scanner;
class ScannerTest
{
public static void main(String[] args)
{
System.out.println("Please give some value");
Scanner sc = new Scanner(System.in);
while([B]sc.hasNextInt[/B]())
{
int i = sc.nextInt();
System.out.println("The value entered is :"+i);
}
sc.close();
}
}
 
 
   

When I compile & run it:

 
C:\sai>javac ScannerTest.java
 
C:\sai>java ScannerTest
Please give some value
12 3 40
The value entered is :12
The value entered is :3
The value entered is :40
 

 

 
   

Eventhough I'm able to get all the values I pass, remember control is not back to the JVM i.e. C:\sai here
It means that hasNextInt() always returns true even after reading all the input values.
How it can be avoided?
Regards.

[ramya narayanan]

When i try to find the end of value in the input stream using

while(sc.nextInt()!=-1)
    

The output I get is:

  
C:\sai>javac ScannerTest.java
 
C:\sai>java ScannerTest
Please give some value
12 34 45
The value entered is :34
 

 

 

 

 
  

Still the challenge gets tougher, not any lesser.
Regards.

[ramya narayanan]

Actually what happens is even after printing the inputstream values , the scanner waits for input from the inputstream .
I've modified my source slightly.

  
 
import java.util.Scanner;
import java.io.*;
class ScannerTest
{
static boolean enable=true;
static int[] inarr=new int[3];
public static void main(String[] args)throws IOException
{
System.out.println("Please give some value");
Scanner sc = new Scanner(System.in);
while(sc.hasNextInt())
{
int i = sc.nextInt();
System.out.println("The value entered is :"+i);
}
 
sc.close();
}
}
 
 
  

output:

  
 
C:\sai>javac ScannerTest.java
 
C:\sai>java ScannerTest
Please give some value
12 34 45
The value entered is :12
The value entered is :34
The value entered is :45
[B]56 78[/B]
[B]The value entered is :56
The value entered is :78[/B]
 
  

See the values 56 & 78 are entered the second time when the scanner is open to get values from the input stream

Now if we want to close the stream
1)we should enforce an constraint on the number of values which can be given by creating an array with some length.

  static int[] inarr=new int[3];
   

2) After the array has been filled with values , close the inputstream .
(Note: If you try to close the scanner you will get IllegalStateException )

if(i==inarr.length)
{
System.in.close();
}
 
    

This is the code

  
import java.util.Scanner;
import java.io.*;
class ScannerTest
{
static boolean enable=true;
[B]static int[] inarr=new int[3];[/B]
public static void main(String[] args)throws [B]IOException[/B]
{
System.out.println("Please give some value");
Scanner sc = new Scanner(System.in);
for(int i=0;i<inarr.length;i++)
{
[B]if(i==inarr.length)
{
System.in.close();
}[/B]
int value=sc.nextInt();
inarr[i]=value;
System.out.println("inarr["+i+"] : "+value);
}
sc.close();
}
}
  

output:

  
C:\sai>javac ScannerTest.java
 
C:\sai>java ScannerTest
Please give some value
12 34 45
inarr[0] : 12
inarr[1] : 34
inarr[2] : 45
 
[B]C:\sai>[/B]
  

Note: The control has been returned.

I want to know is there any way
1) to find the length of the input stream values we give.

2) Can we impose restrictions on the number of values scanner can parse in Scanner or in inputstream on number of values we can give.

Regards.

[Campbell Ritchie]

System.in and System.out and System.err probably don't need to be closed. All the other Streams do, not those three. You probably mean sc.close();

You can have a for loop to count how many values you want from the standard input.
The only way to get -1 from a Scanner reading from System.in is to write -1. It doesn't have a default to -1 for end of stream. But you can write -1 to signify finishing the application if you so wish.

You will occasionally have to use the nextLine() method to move onto a new line, and discard the input.

[ramya narayanan]

System.in and System.out and System.err probably don't need to be closed. All the other Streams do, not those three. You probably mean sc.close();

No, if I try to close sc.close() it throws java.lang.IllegalStateException

You can have a for loop to count how many values you want from the standard input

That's what I've done

The only way to get -1 from a Scanner reading from System.in is to write -1. It doesn't have a default to -1 for end of stream.

I've tried it out by

 
while(sc.hasNextInt==-1)
   

But the problem with this approach
1) It doesn't display all the values I enter through System.in
2) Old problem of non-return of control persists.

You will occasionally have to use the nextLine() method to move onto a new line, and discard the input.

Not understanding this point.
Regards.

[Campbell Ritchie]

Please don't take over somebody else's thread. Please post the questions about Scanner in a new thread; they are not related to the original poster's problem.