I am declaring a String array as:
String[] items = new String[10];
items[0] = "item1";
items[1] = "item2";
How can I find length of items in an efficient way that it contains only 2 elements. items.length returns 10.
I am running a loop already which runs to its length. I want to so something with this code without adding new code/loop to count number of not-null elements. What can I replace with items.length
for (int i = 0; i < items.length; i++) {
...
}
No. You will need to loop and see how many non-null elements there are.
Consider using e.g. an ArrayList<String> instead of a raw array.
UPDATE
To answer the new part of your question, your loop can become:
for (int i = 0; (i < items.length) && (items[i] != null); i++) {
...
}
Why not use a collection:
Vector <String> items;
items.add("item1");
items.add("item2");
int length = items.size();
It is 10 already it is just that 8 of the object are set to null so you could do following
int count = 0 ;
if(items!=null){
for(String str : items){
if(str != null){
count ++;
}
}
}
With the modified question in mind, you can absolutely do nothing. There is no attribute giving you the count of not null elements, so either you'd take another collection or you'd check each value for null.
I think may this will help u
public class T1 {
public static void main(String[] args) {
String[] items = new String[10];
items[0] = "item1";
items[1] = "item2";
System.out.println(getActualSize(items));
}
public static int getActualSize(String[] items)
{
int size=0;
for(int i=0;i<items.length;i++)
{
if(items[i]!=null)
{
size=size+1;
}
}
return size;
}
}
You need to iterate and check for null.
int count = 0;
for(int i = 0; i < items.length; i++){
if(items[i] != null){
count++;
}
}
count will give the number of occupied elements
String[] items = new String[10]; and print all the items in the array. You will see null. So why are you saying there is not null !