java abstract data types?

java abstract data types?
Hi I am writing a program for class using abstract data types,instance variables,constructor and mutator methods. The first part dogs compiles fine the second kennel gives me can not find symbol errors for all instances of dog and any place i have keyboard.nextInt; Thanks for any help!
{CODE}
public class Dogs2
{

private String name;
private String owner;
private String breed;
private int age;

public Dogs2()
{
name=”";
owner=”";
breed=”";
age=0;
}

public Dogs2(String name, String owner, String breed, int age)
{
this.name=name;
this.owner=owner;
this.breed=breed;
this.age=age;
}

public Dogs2(Dogs2 someDog)
{
this.name=someDog.getName();
this.owner=someDog.getOwner();
this.breed=someDog.getBreed();
this.age=someDog.getAge();
}

public String getName()
{
return name;
}
public String getOwner()
{
return owner;
}
public String getBreed()
{
return breed;
}
public int getAge()
{
return age;
}

public int computeYearsToLive()
{
int toLive;
if(breed.equalsIgnoreCase(“german”))
{
toLive=15-age;
}
else if(breed.equalsIgnoreCase(“boxer”))
{
toLive=11-age;
}
else if(breed.equalsIgnoreCase(“terrier”))
{
toLive=8-age;
}
else if(breed.equalsIgnoreCase(“doberman”))
{
toLive=13-age;
}
else
{
toLive=14-age;
}
return toLive;
}
}
{CODE}

{CODE}

import java.util.Scanner;

public class Kennel
{
public static void main( String args[] )
{
Scanner keyboard = new Scanner(System.in);

String Name=”",Owner=”", Breed=”", output=”";
int Age, c, counter;
dog dogArray[] = new dog[10];

for( c=0 ; c <10; c++)
{
Name=keyboard.nextInt.System.out.println("Enter name of dog, type quit when done.");
if(Name.equalsIgnoreCase("quit"))
{
break;
}
Owner=keyboard.nextInt.System.out.println("Enter name of owner of dog");
Breed=keyboard.nextInt.System.out.println("Enter breed of dog");
Age=Integer.parseInt(keyboard.nextInt.System.out.println("Enter age of dog"));
dogArray[c] = new Dogs2(Name, Owner ,Breed, Age);
}
while(true)
{
Name=keyboard.nextInt.System.out.println("Enter name of dog to display, type quit to exit.");
if(Name.equalsIgnoreCase("quit"))
{
break;
}
for( counter =0 ; counter < c; counter++)
{
if(Name.equalsIgnoreCase(dogArray[counter].getName()) )
{
keyboard.nextInt.System.out.println(null, "Your dog info is:\n"
+ "Name: "+dogArray[counter].getName() +
"\nOwner Name: "+dogArray[counter].getOwner() +
"\nBreed: "+dogArray[counter].getBreed()+
"\nAge: "+dogArray[counter].getAge()+
"\nYear Left to Live: "+
dogArray[counter].computeYearsToLive() );
break; //exit for loop
}//end if
}
if (counter==c)
{
keyboard.nextInt.System.out.println(null, "Dog not found! He must have died.");
}
}//end while
keyboard.nextInt.System.out.println(null, "So long!");
System.exit(0);
}
}
[CODE]

Best answer:

It’s quite obvious by your code that you have not been attending class. Pardon my pun, but your code is a bunch of dog shit thrown together. Probably in the hopes that someone will rewrite it all for you. But since I wasted 5 minutes of my life looking at your garbage, the least I could do is give you a starting point on where to go from here…

First of all, you declared Name, Owner, and Breed as strings, yet your miserably failed attempts to receive input for those variables calls the keyboard.nextInt….You can not put an integer into a string obviously. The only one that would work for is age. Second of all, leave the System.out.println’s on their own line. Thirdly, it is good practice to follow standard coding principles…variable names should start with lowercase letters (eg. owner, breed, age)…You have a random “type quit when done”…Well the loop will finish after 10 iterations anyway unless you want to give the user an option to exit before 10 iterations are complete…If that’s the case, than it could be handled more appropriately. I did not even look beyond that code. In the mean time, I have provided a little reworking of that loop. It should hopefully be enough to help you get moving again. But seriously, go back to class…You need it.

for( c=0 ; c < 10; c++)
{
Name=keyboard.next();
System.out.println(Name + " type quit when done.");
if(Name.equalsIgnoreCase("quit")) break;
Owner=keyboard.next();
System.out.println(Owner);
Breed=keyboard.next();
System.out.println(Breed);
Age=keyboard.nextInt();
System.out.println();
dogArray[c] = new Dogs2(Name, Owner, Breed, Age);
}

Tags: abstract, Data, java, types

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