Programming Forums > Application Development > Java

truBlu1 · Sep 29th, 2007, 3:36 AM

This program should be simple. I think, however, I'm misunderstanding how Java handles arrays. This code compiles fine but when my for loop tries to access the elements I get an out of bounds error. My only guess is that I've either declared my array wrong or I've misunderstood how Java handles passing by reference. Please help me.

For reference, the output is supposed to look similar to:
Input: 1, 2, 3, 4
Ouput: 1+4=5, 2+3=5

(Toggle Plain Text)

//Lab6:  Using void methods and 1D arrays to print the sum of numbers in an array (1st + nth, 2nd + nth-1, etc.)
//Input: A bunch of numbers
//Output: A bunch of sums
//Filename: Lab6.java

import java.util.*;

public class Lab6 
{
  public static void main(String [] args)
  {
    System.out.println("How many numbers will be input?");
    
    Scanner keyboard = new Scanner(System.in);
    int n = keyboard.nextInt();
    
    int [] numbers = new int[n];
    getNumbers(numbers);
    
    for(int i = 0; i < n/2; i++)
      System.out.println(numbers[i] + " + " + numbers[n-i] + " = "
                                              + numbers[i]+numbers[(n-1)-i]);
    if(n % 2 != 0)
      System.out.println(numbers[n/2+1] + " could not be added to anything.");
  }
  
  static void getNumbers(int [] numbers)
  {
    System.out.println("Please enter the numbers.");
    Scanner keyboard = new Scanner(System.in);
    for(int i = 0; i < numbers.length; i++)
      numbers[i] = keyboard.nextInt();
    return;
  }
}

Jabo · Sep 29th, 2007, 7:42 AM

Quote:

Originally Posted by truBlu1

(Toggle Plain Text)

for(int i = 0; i < n/2; i++)

i<n/2 is the problem i think. it should be, i<(n/2)-1

DaWei · Sep 29th, 2007, 9:05 AM

First of all, pick some small number like 3 or 5 or something, and go through your code, working it out by hand. After you've done that and fixed your code, then ask yourself what happens if the user enters -12 at the first prompt.

truBlu1 · Sep 29th, 2007, 12:40 PM

Thanks DaWei. I found it immediately when I started going through by hand. I was trying to access numbers[n-i] instead of numbers[(n-1)-i).

I have another question. Look at this segment of code. When my program outputs, it is concatenating the numbers as if they were strings. I get output like 1 + 13 = 113. Why does this happen? I've already fixed the problem by creating an integer variable and setting it equal to numbers[i]+numbers[(n-1)-i] but I'm still curious.

(Toggle Plain Text)

     for(int i = 0; i < n/2; i++)
      System.out.println(numbers[i] + " + " + numbers[(n-1)-i] + " = "
                                              + numbers[i]+numbers[(n-1)-i]);

I also added this to my first prompt:

(Toggle Plain Text)

    int n; //number of numbers to be read into the array
    do
    {
     n = keyboard.nextInt();
     if(n < 0)
       System.out.println("You must input a positive amount of numbers!");
    } while(n < 0);

titaniumdecoy · Sep 29th, 2007, 1:50 PM

Quote:

Originally Posted by truBlu1

I have another question. Look at this segment of code. When my program outputs, it is concatenating the numbers as if they were strings. I get output like 1 + 13 = 113. Why does this happen? I've already fixed the problem by creating an integer variable and setting it equal to numbers[i]+numbers[(n-1)-i] but I'm still curious.

See the documentation for the String class:

Quote:

Originally Posted by Java String Documentation

The Java language provides special support for the string concatenation operator ( + ), and for conversion of other objects to strings. ... String conversions are implemented through the method toString, defined by Object and inherited by all classes in Java.

truBlu1 · Sep 29th, 2007, 2:09 PM

I see. Anything included in println() and used with the + operator is made into a string automatically. I didn't realize that. I thought only what was enclosed in " " was a string.

titaniumdecoy · Sep 29th, 2007, 2:17 PM

No, println() has nothing to do with it.

DaWei · Sep 29th, 2007, 3:27 PM

Actually, println has everything to do with it. Compare the output for these two statements:

(Toggle Plain Text)

    System.out.println (1 + 13);
    System.out.println ("here:" + 1 + 13);

titaniumdecoy · Sep 29th, 2007, 4:53 PM

My point, DaWei, was that the println() method has no effect on how the + operator is treated as the OP suggested in his last post, as this code demonstrates.

(Toggle Plain Text)

Object a = 1 + 13;
Object b = "here:" + 1 + 13;

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

Harakim · Oct 12th, 2007, 10:08 PM

There are three things to be taken from this:
1) PrintStream.println() converts whatever is inside to a string...basically
2) Operations inside a function call's parenthesis are evaulated before the call.
3) Operations in Java are evaluated from left to right.
4) When an operation between two types happens, the lesser type is promoted to the greater type.

The operators in Java are evaluated from left to right:
EXAMPLE: 1 + 2 + 3 + "a" == ((1 + 2) + 3) + "a"
EXAMPLE: 1 + 2 + 3 + "a" != 1 + (2 + (3 + "a"))
More specific examples are included at the bottom

More than you want to know about type promotions
When an operation between two types happens, the lesser type is promoted to the greater type:

(Toggle Plain Text)

EXAMPLE: byte b = (byte)1;
              int a = b + 13434;  //b becomes an int
EXAMPLE: int i = 10;
EXAMPLE: String s = i + "cat"; // i becomes "10", the string

System.out.println( 1 + 2 );
-->
1 + 2;
System.out.println( 3 );

System.out.println( "a" + 1 + 2 );
-->
"a" + 1
"a1" + 2
System.out.println( "a12" );

System.out.println( 1 + 2 + "a" );
-->
1 + 2
3 + "a"
System.out.println( "3a" );

Sep 29th, 2007, 3:36 AM	#1
truBlu1 Newbie Join Date: Sep 2007 Posts: 5 Rep Power: 0	code to add numbers in an array This program should be simple. I think, however, I'm misunderstanding how Java handles arrays. This code compiles fine but when my for loop tries to access the elements I get an out of bounds error. My only guess is that I've either declared my array wrong or I've misunderstood how Java handles passing by reference. Please help me. For reference, the output is supposed to look similar to: Input: 1, 2, 3, 4 Ouput: 1+4=5, 2+3=5 (Toggle Plain Text) //Lab6: Using void methods and 1D arrays to print the sum of numbers in an array (1st + nth, 2nd + nth-1, etc.) //Input: A bunch of numbers //Output: A bunch of sums //Filename: Lab6.java import java.util.; public class Lab6 { public static void main(String [] args) { System.out.println("How many numbers will be input?"); Scanner keyboard = new Scanner(System.in); int n = keyboard.nextInt(); int [] numbers = new int[n]; getNumbers(numbers); for(int i = 0; i < n/2; i++) System.out.println(numbers[i] + " + " + numbers[n-i] + " = " + numbers[i]+numbers[(n-1)-i]); if(n % 2 != 0) System.out.println(numbers[n/2+1] + " could not be added to anything."); } static void getNumbers(int [] numbers) { System.out.println("Please enter the numbers."); Scanner keyboard = new Scanner(System.in); for(int i = 0; i < numbers.length; i++) numbers[i] = keyboard.nextInt(); return; } } //Lab6: Using void methods and 1D arrays to print the sum of numbers in an array (1st + nth, 2nd + nth-1, etc.) //Input: A bunch of numbers //Output: A bunch of sums //Filename: Lab6.java import java.util.; public class Lab6 { public static void main(String [] args) { System.out.println("How many numbers will be input?"); Scanner keyboard = new Scanner(System.in); int n = keyboard.nextInt(); int [] numbers = new int[n]; getNumbers(numbers); for(int i = 0; i < n/2; i++) System.out.println(numbers[i] + " + " + numbers[n-i] + " = " + numbers[i]+numbers[(n-1)-i]); if(n % 2 != 0) System.out.println(numbers[n/2+1] + " could not be added to anything."); } static void getNumbers(int [] numbers) { System.out.println("Please enter the numbers."); Scanner keyboard = new Scanner(System.in); for(int i = 0; i < numbers.length; i++) numbers[i] = keyboard.nextInt(); return; } }

Sep 29th, 2007, 9:05 AM	#3
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	First of all, pick some small number like 3 or 5 or something, and go through your code, working it out by hand. After you've done that and fixed your code, then ask yourself what happens if the user enters -12 at the first prompt. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Sep 29th, 2007, 12:40 PM	#4
truBlu1 Newbie Join Date: Sep 2007 Posts: 5 Rep Power: 0	Thanks DaWei. I found it immediately when I started going through by hand. I was trying to access numbers[n-i] instead of numbers[(n-1)-i). I have another question. Look at this segment of code. When my program outputs, it is concatenating the numbers as if they were strings. I get output like 1 + 13 = 113. Why does this happen? I've already fixed the problem by creating an integer variable and setting it equal to numbers[i]+numbers[(n-1)-i] but I'm still curious. (Toggle Plain Text) for(int i = 0; i < n/2; i++) System.out.println(numbers[i] + " + " + numbers[(n-1)-i] + " = " + numbers[i]+numbers[(n-1)-i]); for(int i = 0; i < n/2; i++) System.out.println(numbers[i] + " + " + numbers[(n-1)-i] + " = " + numbers[i]+numbers[(n-1)-i]); I also added this to my first prompt: (Toggle Plain Text) int n; //number of numbers to be read into the array do { n = keyboard.nextInt(); if(n < 0) System.out.println("You must input a positive amount of numbers!"); } while(n < 0); int n; //number of numbers to be read into the array do { n = keyboard.nextInt(); if(n < 0) System.out.println("You must input a positive amount of numbers!"); } while(n < 0);

Sep 29th, 2007, 3:27 PM	#8
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	Actually, println has everything to do with it. Compare the output for these two statements: (Toggle Plain Text) System.out.println (1 + 13); System.out.println ("here:" + 1 + 13); System.out.println (1 + 13); System.out.println ("here:" + 1 + 13); __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Sep 29th, 2007, 4:53 PM	#9
titaniumdecoy Expert Programmer Join Date: Nov 2005 Posts: 836 Rep Power: 3	My point, DaWei, was that the println() method has no effect on how the + operator is treated as the OP suggested in his last post, as this code demonstrates. (Toggle Plain Text) Object a = 1 + 13; Object b = "here:" + 1 + 13; System.out.println(a); System.out.println(b); Object a = 1 + 13; Object b = "here:" + 1 + 13; System.out.println(a); System.out.println(b);

Sep 29th, 2007, 2:09 PM	#6
truBlu1 Newbie Join Date: Sep 2007 Posts: 5 Rep Power: 0	I see. Anything included in println() and used with the + operator is made into a string automatically. I didn't realize that. I thought only what was enclosed in " " was a string.

Sep 29th, 2007, 2:17 PM	#7
titaniumdecoy Expert Programmer Join Date: Nov 2005 Posts: 836 Rep Power: 3	No, println() has nothing to do with it.

Oct 12th, 2007, 10:08 PM	#10
Harakim Hobbyist Programmer Join Date: May 2006 Location: West Jordan, Utah, United States Posts: 176 Rep Power: 3	There are three things to be taken from this: 1) PrintStream.println() converts whatever is inside to a string...basically 2) Operations inside a function call's parenthesis are evaulated before the call. 3) Operations in Java are evaluated from left to right. 4) When an operation between two types happens, the lesser type is promoted to the greater type. The operators in Java are evaluated from left to right: EXAMPLE: 1 + 2 + 3 + "a" == ((1 + 2) + 3) + "a" EXAMPLE: 1 + 2 + 3 + "a" != 1 + (2 + (3 + "a")) More specific examples are included at the bottom More than you want to know about type promotions When an operation between two types happens, the lesser type is promoted to the greater type: (Toggle Plain Text) EXAMPLE: byte b = (byte)1; int a = b + 13434; //b becomes an int EXAMPLE: int i = 10; EXAMPLE: String s = i + "cat"; // i becomes "10", the string EXAMPLE: byte b = (byte)1; int a = b + 13434; //b becomes an int EXAMPLE: int i = 10; EXAMPLE: String s = i + "cat"; // i becomes "10", the string System.out.println( 1 + 2 ); --> 1 + 2; System.out.println( 3 ); System.out.println( "a" + 1 + 2 ); --> "a" + 1 "a1" + 2 System.out.println( "a12" ); System.out.println( 1 + 2 + "a" ); --> 1 + 2 3 + "a" System.out.println( "3a" );

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Programming Forums > Application Development > Java

code to add numbers in an array