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Thread: BigElement-HELP
- 03-05-2009, 05:16 PM #1
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BigElement-HELP
Assignment number 2:
Create a program called BigElement that will allow the user to input five integers. The program will then output the largest of the five numbers. Include a function-type method that will return to the main method the value of the largest element in the array (Hint: the flowchart below shows the algorithm that could be used for the method). Use the length method to help determine the number of elements in the array.
This is assignment 2 that I need to complete quickly and all I have is this:
// The "BiggestElement" class.
import java.awt.*;
import hsa.Console;
public class BiggestElement
{
static Console c; // The output console
public static void main (String[] args)
{
c = new Console ();
int biggest[] = new int [5];
//i know this part is wrong, im not sure what to do exactly
c.println ("Please input five integers: ");
biggest = c.readInt[] ();
} // main method
} // BiggestElement class
- 03-05-2009, 07:01 PM #2
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if you have 5 elements, make a "for" loop, initialize a int variable max=0
in each time, you compare max with each element, if max is smaller, than assign max as that element,
hope it helps
- 03-06-2009, 02:16 AM #3
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Sorry not quite sure I got it....would you happen to have msn so I can ask a few questions quicker? ...I aprreciate the help.
c = new Console ();
int biggest[] = new int [5];
c.println ("Please input five integers: ");
biggest = c.readInt ();
for (int max = 0 ; max < biggest ; max++)
{
}
} // main method
- 03-06-2009, 02:49 AM #4
pseudo code
It would llook something like this:
Luck,Java Code:create array of 5 elements start loop ask user for int assign int to array end loop call method (sortMethod?) with array as a parameter print the int retunred from the method begining of sort method do sort stuff return the biggest element (int) end method
CJSLChris S.
Difficult? This is Mission Impossible, not Mission Difficult. Difficult should be easy.
- 03-06-2009, 04:47 PM #5
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Ok so this is what I have now, im not sure if my loop in the sortMethod is correct, also i get an error when i try to print the method out from the main method
Java Code:public static void main (String[] args) { c = new Console (); //create array of 5 elements int fiveNumber[] = new int [5]; int max = 0; //ask the user for five integers c.println ("Please input five integers: "); //start a loop so that the user can enter five integers for (int i = 0 ; i < 5 ; i++) { fiveNumber [i] = c.readInt (); } c.println ("The largest number entered was " + sortMethod (max)); } // main method public static int sortMethod (int fiveNumber, int max) { for (int i = 0 ; i < 5 ; i++) { if (fiveNumber > max) max = fiveNumber; } return max; }
- 03-06-2009, 05:18 PM #6
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nevermind i got the program to run!:
Java Code:c = new Console (); //create array of 5 elements int fiveNumber[] = new int [5]; //declare a variable to determine the largest number int max = 0; //ask the user for five integers c.println ("Please input five integers: "); //start a loop so that the user can enter five integers for (int i = 0 ; i < 5 ; i++) { fiveNumber [i] = c.readInt (); } //print the largest number, that has been returned from the sortMethod method c.println ("The largest number entered was " + sortMethod (fiveNumber, max)); } // main method //create a new method called SorthMethod to determine which of the five numbers is the largest public static int sortMethod (int[] fiveNumber, int max) { //use a loop to determine out of five numbers which is the largest for (int i = 0 ; i < 5 ; i++) { if (fiveNumber [i] > max) max = fiveNumber [i]; } //return the max integer, which is now the largest integer in the given array return max; }
- 03-06-2009, 06:05 PM #7
Great... glad it works !!!
a comment... there's no need to pass variable max as a parameter (it's value is zero at that point). Just declare it and initialize it in the sort method.Java Code:public static int sortMethod (int[] fiveNumber, [B][COLOR="Red"]int max[/COLOR][/B])
Luck,
CJSLChris S.
Difficult? This is Mission Impossible, not Mission Difficult. Difficult should be easy.


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