Hey all! My program is supposed to take an array of 10 integers and shift them by as many as the user wants (which can be up to 9 because of you shift the 10th integer in the array you would get the same array you started with back).

So lets say this is your array: 1,2,3,4, 5, 6, 7, 8, 9,10

And the user enters 3 as the number to shift by: then you should get an output of 8, 9, 10, 1, 2, 3, 4, 5, 6, 7

Here is my code:
import javax.swing.JOptionPane;
public class shiftarray {
public static void main(String[] args) {
int[] a = new int[10];
int[] b = new int[10];
for (int j = 0; j < a.length; j++) {
int n = Integer.parseInt(JOptionPane.showInputDialog(null,
"Enter an integer: "));
}
for (int i=0; i < n; i++){
b[i+1] = a[i];
if(a[i+1] > a.length-1)
b[0]=a[i];
}
}
}

It doesn't work. It runs and lets you enter a number but it stops after you enter a number. Any ideas as to why this is and what I need to do to get this program running correctly?

 

 

b[i+1] = a;


how does this even compile?

 

 

Also, the variable 'n' is out of scope on your last for loop.
It doesn't compile, nor does it execute. You think you posted the wrong code?

 

 

Please put your code between code tags.

How are you filling the array? Does it even compiles? (The code you provided will not compile BTW) Are you getting an error?

 

 

It should be

b[i+1] = a[i];

Due to the unformatted code the i is treated as the italics tag.

 

 

Assuming that you take out and initialize 'n' so that it is in the scope of the last for loop.

You still have a major chance of the the second for loop going out of bounds.
For instance, if the last value in 'n' was 200; then the for loop runs until i < 200.

And of course, you're array is not 200 slots big.

Now, if you replace the second for loop with

for(int i = 0; i < a.length; i++)

you will see you will still get OutOfBounds errors.
For example, when i becomes 9, the next statement:

b[i+1] = a[i];

is the same as

b[10] = a[9]

And by the given code, the highest spot for b is b[9]

 

 

Also, you're shifting the two arrays without initializing them, so they're basically {0,0,0,0,0,0,0,0,0,0}
It doesn't matter how much you rotate them, they'll always be the same ;-)

 

 

Yes, there is obviously quite a few errors in the given code.

I think in the first for loop you mean to initialize the array with the values of n.
In that case you would need:

for(int i = 0; i < a.length; i++) {
    n = Integer.parseInt(JOptionPane.showInputDialog(
                                         null, "Enter an Integer");
 
    a[i] = n;
}

null

 

 

From what I understand of the first question, the initialisation loop should be:

for (int i=0; i < a.length; ++i) {
    a[i] = i+1;
} 

your shifting loop should not loop from 0 to n, cause that will only shift n elements 1 spot. You have to loop from 0 to a.length-1 and set

b[ i ] = a[(i + n) % a.length]

It might also be a good idea to print something after you shift the values, otherwise it might look like the code just stops...

Message was edited by:
Peetzore

 

 

The simple logic here goes:
if array size is 10 and n is 3,
1. Start from 7 th position. Move 7th element to 0th of new array, 8th to 1st and so on.
Now you got shifted elements in base position of new array.
2. Now take the 1st element of source array and move it to n'th position of new array, 2nd to n+1 etc.

Does it sound clear ?

 

 

b[ i ] = a[(i + n) % a.length]

Of course this should be

b[(i + n) % a.length] = a[ i ]

 

 

You can even do that rotate in place, i.e. without a additional array. All
that is needed is a 'reverse' method that reverses (part of) an array:

int[] reverse(int[] a, int i, int j) { // reverse elements i, i+1 ... j-1
   for (; --j > i; i++) {
      int t= a[i]; a[i]= a[j]; a[j]= t;
   }
   return a; // convenience.
}

Here's the shiftRight method in terms of the reverse method:

int[] shiftRight(int[] a, int i) {
   return reverse(reverse(reverse(a, 0, a.length), 0, i), i, a.length);
}

I'm sure you can do a shiftLeft method yourself now.

kind regards,

Jos

 

 

Nice one, Jos!

 

 

Thanks; it's an ancient trick from the times when memory was measured
in <everybody goes: aaahhhhh!>Kilo bytes</aaahhh> You can move
an arbitrary block around an array without using an additional array.
It's mostly forgotten nowadays ;-)

kind regards,

Jos

 

 

Ahh, the good ol' days... Wasn't that the time when animals used to talk? Guess I'm one of those young luxorious little puppies that can't imagine what that time could have been like!