// Teddybot
// Random, Shuffling, Arrays, Passing array pointer

void setup() {
  println("Create array 1,2,3,4,5...");
 int[] a = new int[5];
 for(int i=0; i<a.length; i++)
     a[i] = i+1;
 display(a); //displays 1 2 3 4 5

 shuffle(a);
 display(a);

 randomize(a);
 display(a);

}

void randomize(int[] a)
{
  println("Randomizing...");
   for(int i=0; i<a.length; i++)
     a[i] = (int)random(5)+1;
}

void shuffle(int[] a)
{
  int temp;
  int pick;
 
  println("Shuffling...");
   for(int i=0; i<a.length; i++)
     {
       temp = a[i];
       pick  = (int)random(5);
       a[i] = a[pick];
       a[pick]= temp;
     }
}

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

teddybot,

By randomise the order I assume you mean to shuffle them.

I have modified the shuffle code in the last post to work with an array of any size to make sure that

• each position gets swapped (if you assumed array was smaller than actual size)
• avoids the ArrayOutOfBounds exception (if you assumed the array was longer than actual size)

1. void suffle(int[]value){
2.    int temp;
   
3.    for(int i=0;i<value.length;i++)
4.        pick  = (int)random(value.length); // picks a ronaodm position in the array
   
5.        temp = value[i]; // stores value of current position
   
6.        value[i] = a[pick]; // copies picked value into current position
   
7.        value[pick]= temp; // store original value in picked position
   
8.     }
   
9. }

  

You know, it is among the most frequently asked questions... Have you tried to search the forum? You can add the "shuffle" word that have been thrown above. There are several solutions, including a one liner with Java collections.