Processing 1.0 - Processing Discourse - Basic Question: ArrayIndexOutOfBoundsException: 0

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Basic Question: ArrayIndexOutOfBoundsException: 0 (Read 742 times)

faebser

Basic Question: ArrayIndexOutOfBoundsException: 0
Oct 29^th, 2009, 3:22am

Hi there,

i'm currently around with arrays, coming from a php-background.
so there is alot different in handling array's in processing then in php.
i did a rather small thing to learn arrays but i always get an ArrayIndexOutOfBoundsException: 0 on line 10.
can anybody explain me what i'm doing wrong?
Code:

int num;
int[] ints;

num = 0;
ints = new int[num];
println(ints.length);

for(int i = 0;i <= 50;i++)
{
  ints[i] = int(random(0,100));
  println(ints[i]);
}

nardove

Re: Basic Question: ArrayIndexOutOfBoundsException: 0
Reply #1 - Oct 29^th, 2009, 3:30am

Hi, someone correct me if there is better way to solve this

First I think you need to initialize the array with the number of items you will use in the for loop

Second the for loop cant look the array if the condition is set to <= because it will loop 51 times

Code:

int num;
int[] ints;

num = 50;
ints = new int[num];
println(ints.length);

for(int i = 0; i < num;i++)
{
  ints[i] = int(random(0,100));
  println(ints[i]);
}

Have a look in to arrayList as well, they are more "dynamic"

Cheers
rS

blindfish

Re: Basic Question: ArrayIndexOutOfBoundsException: 0
Reply #2 - Oct 29^th, 2009, 3:42am

Exactly as nardove said. The '<=' is why you're getting an outOfBounds error since it will try and access ints[50] which, since array indices start at 0, doesn't exist.

You can also use shortcuts on variable declarations and should avoid the use of 'magic numbers':

Code:

int arrayLength = 50;
int[] ints = new int[arrayLength];
println("length: " + ints.length);

// Either use arrayLength or ints.length as the limit:
for(int i = 0;i < arrayLength; i++) {
  ints[i] = int(random(0,100));
  println(i + ": " + ints[i]);
}

Where is the what if the what is in why?

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