reverse for loop for array countdown
Solution 1
Java uses 0-based array indexes. When you create an Array of size 10 new int[10]
it creates 10 integer 'cells' in the array. The indexes are: 0, 1, 2, ...., 8, 9.
Your loop counts to the index which is 1 less than 11, or 10, and that index does not exist.
Solution 2
You declared array on integers of 10 elements
. And you are iterating from i=0 to i=10
and i=10 to i=0
that's 11 elements
. Obviously it's an index out of bounds error
.
Change your code to this
public class Reverse {
public static void main(String [] args){
int i, j;
System.out.print("Countdown\n");
int[] numIndex = new int[10]; // array with 10 elements.
for (i = 0; i<10 ; i++) { // from 0 to 9
numIndex[i] = i;// element i = number of iterations (index 0=0, 1=1, ect.)
}
for (j=9; j>=0; j--){ // from 9 to 0
System.out.println(numIndex[j]);//indexes should print in reverse order from here but it throws an exception?
}
}
}
Remember indices starts from 0.
.
Solution 3
The array is of size 10, which means it is indexable from 0 to 9. numIndex[10]
is indeed out of bounds. This is a basic off-by-one error.
Solution 4
An Array
in java that has 10
elements goes from 0
to 9
. So your loops need to cover this range. Currently you are going from 0
to 10
, and 10
to 0
.
KamikazeStyle
Updated on March 22, 2020Comments
-
KamikazeStyle about 4 years
I get the error..
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 10 at Reverse.main(Reverse.java:20).
There is not wrong in the syntax so im not sure why when it compiles it gets an error?
public class Reverse { public static void main(String [] args){ int i, j; System.out.print("Countdown\n"); int[] numIndex = new int[10]; // array with 10 elements. for (i = 0; i<11 ; i++) { numIndex[i] = i;// element i = number of iterations (index 0=0, 1=1, ect.) } for (j=10; j>=0; j--){ // could have used i, doesn't matter. System.out.println(numIndex[j]);//indexes should print in reverse order from here but it throws an exception? } }
}