+ 1
ARRAYS JAVA
public class MyClass { public static void main(String[ ] args) { int a[ ] = new int[5]; System.out.println(a [1] ); } } Why output "0"????And if we change number to "5" - we will have error. Why?? Give me pls advise where I can get more information about ARRAYS?
3 Answers
+ 4
When you create an integer array, by default, java places 0 at each index of the array. You can find more information at official java documentation https://docs.oracle.com/javase/tutorial/java/nutsandbolts/arrays.html
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a[5] is beyond the end of the array.
When you create an array like:
int a[] = new int[5];
this creates an array with 5 elements. Arrays are zero based, however, and the first element is at a[0] not a[1], making the last element a[4].
0
First of all You should understand difference between index of array and length of array.
index of array is start with 0
length of array is start with 1
So If you store the 5 elements to the array your length of array will be 5
the number which is inside of the square brackets represent length of the array
If you want to get element from the array you will use index of array.
For instance,
1- Apple
2- Orange
3- Banana
if you want to print Apple, System.out.println(String[1-1]);
if you want to print Orange System.out.println(String[2-1]);
if you want to print Banana, System.out.println(String[3-1]);
So,
index of 0 will be first element
index of 1 will be second element ......
index of 4 will be last element in the array