Simple Java Array Question

Question

// This is how I was taught to make a simple array. You have to define the size first.
String [] test1 = new String[4];
test1[0] = "1";
test1[1] = "2";
test1[2] = "3";
test1[3] = "4";
out.println(test1[0]);
out.println(test1[1]);
out.println(test1[2]);
out.println(test1[3]);
//If the size is not defined in this second example, why does it still work?
String [] test2 = {"1","2","3","4"};
out.println(test2[0]);
out.println(test2[1]);
out.println(test2[2]);
out.println(test2[3]);
They both produce the same results.

Answer 1

// This is how I was taught to make a simple array.
You have to define the size first.
String [] test1 = new String[4];
test1[0] = "1";
test1[1] = "2";
test1[2] = "3";
test1[3] = "4";
out.println(test1[0]);
out.println(test1[1]);
out.println(test1[2]);
out.println(test1[3]);
//If the size is not defined in this second example,
why does it still work?
String [] test2 = {"1","2","3","4"};
out.println(test2[0]);
out.println(test2[1]);
out.println(test2[2]);
out.println(test2[3]);
They both produce the same results.Umm..the size is just as well defined in your second as in your first...it is a four element array.
The second way is akin to the first...also
String[] s = new String[]{"Hello","I","contain","five","elements"};
System.out.println(s[5]); Will compile..you will get a run-time ArrayIndexOutOfBoundsException (it is a 5 element array, 0-4, but java will not assume that the variable s is not changed between those two lines of code)..
Array index/boundary checking is done at run-time, not compile-time.
In fact:
String[] s;
//let's say up in the program
//and
public void printS(int i)
System.out.println(s);
Will also compile...and if something else happens to set s as a 10 element String array:
printS(9);would work
~David