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Java problem

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  • 11-06-2014 5:30pm
    #1
    whats newxt
    Closed Accounts Posts: 431 ✭✭


    //So i'm just learning java myself for the last few days and i have a basic program:

    import java.util.Scanner;

    public class test1 {
    public static void main(String args[]){
    int a = 10;
    int b = 20;

    Scanner v1 = new Scanner(System.in);
    System.out.println("Number for a is: " + a + v1.nextLine());

    Scanner v2 = new Scanner(System.in);
    System.out.println("Number for b is: " + b + v2.nextLine());

    System.out.println(a);
    System.out.println(b);

    }
    }

    //the program works but not as i intended it to, i want to have two integers a and b and i want to take user input (Scanner) and add the user input to the integer store it and print the stored value of a and b at the end. What am i doing wrong. Thanks for your help
    Tagged:


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    #2
    Maximus Alexander
    Registered Users Posts: 13,080 ✭✭✭✭Maximus Alexander


    Both integers a and b are set to 0, you're printing those zero values along with the any input from the scanner. You never actually assign values to a or b.

    Change
    System.out.println("Number for a is: " + a + v1.nextLine());

    to
    a = v1.nextInt();
System.out.println("Number for a is: " + a);


  • Options
    #3
    whats newxt
    Closed Accounts Posts: 431 ✭✭whats newxt


    Maximus Alexander wrote: »
    Both integers a and b are set to 0, you're printing those zero values along with the any input from the scanner. You never actually assign values to a or b.

    Change
    System.out.println("Number for a is: " + a + v1.nextLine());

    to
    a = v1.nextInt();
System.out.println("Number for a is: " + a);


    Thanks it worked: 
    import java.util.Scanner;

public class test1 {
	public static void main(String args[]){
		int a = 10;
		int b = 20;
		
		Scanner v1 = new Scanner(System.in);
		a = a + v1.nextInt();
		System.out.println("Number for a is: " + a);
		
		Scanner v2 = new Scanner(System.in);
		b = b + v1.nextInt();
		System.out.println("Number for b is: " + b);
		
		System.out.println("The stored value foa a is: " + a);
		System.out.println("The stored value for b is: " + b);

	}
}
    [/QUOTE]


  • Options
    #4
    Maximus Alexander
    Registered Users Posts: 13,080 ✭✭✭✭Maximus Alexander


    whats newxt wrote: »
    Thanks it worked: 
    import java.util.Scanner;

public class test1 {
	public static void main(String args[]){
		int a = 10;
		int b = 20;
		
		Scanner v1 = new Scanner(System.in);
		a = a + v1.nextInt();
		System.out.println("Number for a is: " + a);
		
		Scanner v2 = new Scanner(System.in);
		b = b + v1.nextInt();
		System.out.println("Number for b is: " + b);
		
		System.out.println("The stored value foa a is: " + a);
		System.out.println("The stored value for b is: " + b);

	}
}

    The important thing is though, do you understand why it worked? :) Because I just noticed that I misread your original intention.


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    #5
    whats newxt
    Closed Accounts Posts: 431 ✭✭whats newxt


    I'm not to sure i think it's because when you use .nextLine it thinks you're working with a string and when your use .nextInt it knows you're working with integers?


  • Options
    #6
    Maximus Alexander
    Registered Users Posts: 13,080 ✭✭✭✭Maximus Alexander


    Yes that's true, and more specifically, the + operator has two functions. On numeric types it performs addition, on Strings it performs concatenation.

    Because you originally had your addition (a + the value of nextLine()) as a parameter to the System.out.println() function, which takes a String, it just concatenated the string representation of a with whatever came in via the scanner. This would still have happened if you had used nextInt().

    For example if:

    a = 10
    v1.nextInt() = 10

    You'd have gotten 1010.


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  • Options
    #7
    whats newxt
    Closed Accounts Posts: 431 ✭✭whats newxt


    Maximus Alexander wrote: »
    Yes that's true, and more specifically, the + operator has two functions. On numeric types it performs addition, on Strings it performs concatenation.

    Because you originally had your addition (a + the value of nextLine()) as a parameter to the System.out.println() function, which takes a String, it just concatenated the string representation of a with whatever came in via the scanner. This would still have happened if you had used nextInt().

    For example if:

    a = 10
    v1.nextInt() = 10

    You'd have gotten 1010.

    Thanks i see what was happening now, can you recommend any books for learning java most the tutorials on you tube skip over important parts.


  • Options
    #8
    Maximus Alexander
    Registered Users Posts: 13,080 ✭✭✭✭Maximus Alexander


    I got the older 8th edition of this a few years ago and I would highly recommend it. Starts with the basics and works up, covers a lot of ground. It's a big book but it's quite comprehensive and covers the details well. It also has plenty of exercises.


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