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I've had a decent search and am unable to find working code that moves down an array. What I am hoping to do, is to store the value in the last position in the array, replace the last position and then move array[20] to array[19]. This is meant to count the last 20 moves the player makes, but I'm having trouble actually storing. This is what I have attempted to do

//an int moveArray[20] previously stated and instantiated
int temp1, temp2;

for (int i = moveArray.length - 1; i > 0; i--) 
{
     temp1 = moveArray[i - 1];
     temp2 = moveArray[i - 2];
     moveArray[i - 1] = moveArray[i];
     temp1 = temp2;
}
moveArray[moveArray.length - 1] = intoWalk;

any advice or solutions would really help, thanks

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2
  • Your question is not clear. If you have an array with 20 elements, you want to store the 20th element in the 19th and also store the value contained in the 19th element prior to replace. What exactly is it that you plan to do with the stored value ? Do you want to move it to 18 ? Commented May 22, 2013 at 5:24
  • count the last 20 moves the player makes - sounds like you need a deque (FIFO) of size 20. Commented May 22, 2013 at 5:43

4 Answers 4

Reset to default
6

From what I understand of your code . You should use the following loop, there seems to be no need for temporary variables.

for(int i=0;i<moveArray.length-1;i++){ 
     moveArray[i] = moveArray[i+1];
}
moveArray[moveArray.length - 1] = intoWalk;
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3 Comments

Thank you very much, this works perfectly and I never even realised I could do it without temps, Thanks again!
I don't know why this was downvoted. It's exactly what I would do and it's a good answer.
Even then you should also read the answer by Florent regarding Linked Lists,it can be used when you don't know the no. of past states to be stored.
3

First, you don't need to use temporary variables :

int [] moveArray = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20};
int intoWalk = 21;

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

for (int i=0; i<moveArray.length; i++)
    System.out.println(moveArray[i]);

But there is a better way to do it : use a linked list to emulate a FIFO :

LinkedList<Integer> fifo = new LinkedList<Integer>();
int intoWalk = 21;

for (int i=1; i<=20; i++)
    fifo.add(i);

fifo.removeFirst();
fifo.add(intoWalk);

for (Integer fifoItem : fifo)
    System.out.println(fifoItem);

By doing this, you don't have to modify every element in your array each time you want to add a number, just to add and remove an item from the linked list.

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Comments

0 
//an int moveArray[20] previously stated and instantiated
int temp1, temp2;
temp1 = moveArray[moveArray.length - 1];
for (int i = moveArray.length - 2; i >= 0; i--) 
{
    temp2 = moveArray[i];
    moveArray[i] = temp1;
    temp1 = temp2;
}

moveArray[moveArray.length - 1] = intoWalk;

The above should do what you want.

However, I don't think this is the right way to go.

I would use the same array as a circular array - that way - you don't need to move down everytime.

Maintain a start index. It begins at

start = 0;

It remains at 0 till the time all 20 elements are filled up. When the next element comes in,

moveArray[start] = intoWalk;
++start;

When you want to iterate through the array at any time, you begin at start instead of beginning at 0.

int i = start;
do
{
    // do what you want

    i = (i + 1)%20; 

} while (i != start);
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Comments

0

First make the last element free by moving all the elements one row down.

Sample :

String[] sarr=new String[10];
for(int i=0;i<sarr.length;i++){
   sarr[i]=sarr[i+1];
}

Then assign the new variable to the end.

sarr[sarr.length-1]="new value";

This approach is simpler and more readable.

EDIT : Yes Bohemian is right it was throwing an ArrayIndexOutOfBounds exception. The reason was that i+1 will try to refer to an index out of the array's range at the last iteration.

The solution is either to set the last element to null or break when it's the last element.

I will go with setting the last element to null. The changed code is below.

int nextElementIndex = i + 1;
sarr[i] = nextElementIndex < sarr.length ? sarr[nextElementIndex] : null;
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6 Comments

This code will throw an index out of bounds exceptopn. Try i < arr.length - 1. Also it's int[] not String[]. Otherwise good answer
@Bohemian i < arr.length - 1 was not the problem. But you are right there's an error. I corrected it.
And also I just wanted to show OP a simple sample rather than changing his code which in my opinion was overcomplicated. It's pretty easy to change it to any type.
Yes I know I corrected it in the edit. I was just telling you that it wasn't because of the loop condition :)
Apparently you don't know, because right now the code you have posted to move the elements down will explode! Ill spell it out for you: In the last iteration, i will be sarr.length - 1, which is the index of the last element, but your code references sarr[i + 1]... BOOM!
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