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Thread: nextint() Method !

  1. #1
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    Angry nextint() Method !

    hi all

    i need help in this exercise :-

    develop main method that initializes a random object with the default constructor and then determines the elapsed time for the nextint() method to generate 123456789

    -------

    how


  2. #2
    Super Moderator copeg's Avatar
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    Default Re: nextint() Method !

    Define help...what do you have so far? If nothing, start here: The Java™ Tutorials

  3. #3
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    Default Re: nextint() Method !

    Quote Originally Posted by copeg View Post
    Define help...what do you have so far? If nothing, start here: The Java™ Tutorials


    this is my basic idea

     Random r = new Random();
     
     for(int i=1;i<10;i++){
     int Number = r.nextInt(10);
     System.out.println(Number);

    this code generate 9 random numbers , now i need to generate 123456789 using this method

    anyone help me

  4. #4
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    Default Re: nextint() Method !

    ok i trying in this idea :-


    Random r = new Random();
    for(int i=0;i<9;i++){
         int[] array=new int[9];
     int Number = r.nextInt(10);
     if((Number==1) || (Number==2) || (Number==3)||(Number==4)||(Number==5)||(Number==6)||(Number==7)||(Number==8)||(Number==9))
             Number=array[i];
     System.out.println(array[i]);
        }


    the output is :

    0
    0
    0
    0
    0
    0
    0
    0
    0


    Why !! why it not working

  5. #5
    Super Moderator copeg's Avatar
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    Default Re: nextint() Method !

    When equating two values, the left operand is assigned the value of the right. Inspect your code carefully and you will see that the values within the array are never assigned anything (and thus you print their default value of 0)

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    M7MD (October 24th, 2010)

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    Member Darryl.Burke's Avatar
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    Default Re: nextint() Method !


  8. #7
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    Default Re: nextint() Method !

    ???????????

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    Default Re: nextint() Method !

    ???????????
    !!!!!!!!!!!

  10. #9
    Super Moderator helloworld922's Avatar
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    Default Re: nextint() Method !

    Quote Originally Posted by M7MD View Post
     int Number = r.nextInt(10);
     if((Number==1) || (Number==2) || (Number==3)||(Number==4)||(Number==5)||(Number==6)||(Number==7)||(Number==8)||(Number==9))
    That condition is completely unnecessary. The definition of nextInt() is that it will return an integer in the range [0,10). Simply change your random-number generation statement to this and there's no need for an if statement.

    int number = r.nextInt(9) + 1;
    // no need for an if statement, number is guaranteed to be in the range [1,10]
    array[i] = number;
    // ...other code

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    M7MD (October 27th, 2010)

  12. #10
    mmm.. coffee JavaPF's Avatar
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    Default Re: nextint() Method !

    I would do it like this:

    import java.util.Random;
     
    public class Numbers {
     
    	/**
    	 * JavaProgrammingForums.com
    	 */	
    	public static void main(String[] args) {
     
    		Random r = new Random();
     
    		int Number;
    		int Count = 1;
     
    		while(Count != 10){
     
    			Number = r.nextInt(10)+1;
     
    			if(Number == Count){
    				System.out.print(Number + " ");
    				Count++;
    			}						
    		}
    	}	 
    }

    Output:

    1 2 3 4 5 6 7 8 9
    Please use [highlight=Java] code [/highlight] tags when posting your code.
    Forum Tip: Add to peoples reputation by clicking the button on their useful posts.

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  13. The Following User Says Thank You to JavaPF For This Useful Post:

    M7MD (October 27th, 2010)

  14. #11
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    Default Re: nextint() Method !

    Quote Originally Posted by JavaPF View Post
    I would do it like this:

    import java.util.Random;
     
    public class Numbers {
     
    	/**
    	 * JavaProgrammingForums.com
    	 */	
    	public static void main(String[] args) {
     
    		Random r = new Random();
     
    		int Number;
    		int Count = 1;
     
    		while(Count != 10){
     
    			Number = r.nextInt(10)+1;
     
    			if(Number == Count){
    				System.out.print(Number + " ");
    				Count++;
    			}						
    		}
    	}	 
    }

    Output:
    Man !

    thanks very much

    100%

    and i'am understand the idea in true way

  15. #12
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    Default Re: nextint() Method !

    Quote Originally Posted by helloworld922 View Post
    That condition is completely unnecessary. The definition of nextInt() is that it will return an integer in the range [0,10). Simply change your random-number generation statement to this and there's no need for an if statement.

    int number = r.nextInt(9) + 1;
    // no need for an if statement, number is guaranteed to be in the range [1,10]
    array[i] = number;
    // ...other code
    Yes !

    thanks for your useful idea

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