This is the snippet of Java code.
class Test{
public static void main(String[ ] args){
int[] a = { 1, 2, 3, 4 };
int[] b = { 2, 3, 1, 0 };
System.out.println( a [ (a = b)[3] ] );
}
}
Why does it print 1? This is not a homework! I am trying to understand Java. That is related to OCA Java 7 exam.
At the moment you refer to a[ ... ], a still points to the first array. When the index itself is evaluated, there's an assignment of b to a. So at that moment, a becomes b, of which the 3rd item is fetched, which is 0.
This 0 is used as an index of the array that was already found before. This is the array that a pointed to, although a itself in the mean time has changed. Therefor it prints the 1, even though you might expect 2.
I think that is what this example is trying to show: The array reference is already evaluated and doesn't change once you modify the array variable during the evaluation of the index.
But I wouldn't use this 'feature' in production code. Very unclear.
System.out.println( a [ (a = b)[3] ] );
First, the value of a is evaluated ({1, 2, 3, 4}). Next, a = b is executed; this assigns the value of b to a and also returns the value of b. b[3] = { 2, 3, 1, 0 } is 0, so, ultimately, {1,2,3,4}[b[3]] = {1,2,3,4}[0] = 1.
To see this, consider the following:
public static void main(String[] args) throws FileNotFoundException {
int[] a = { 1, 2, 3, 4 };
System.out.println( a() [ (a = b())[c()] ] );
}
public static int[] a() {
System.out.println('a');
return new int[]{ 1, 2, 3, 4 };
}
public static int[] b() {
System.out.println('b');
return new int[]{ 2, 3, 1, 0 };
}
public static int c() {
System.out.println('c');
return 3;
}
Output:
a
b
c
1
a[x] is executed first, but x = b[3], so a[b[3]] is 1.
2, not 1.(a = b) not only returns { 2, 3, 1, 0 }, but also assigns that array to a, in which case a[0] would return 2. Therefor, a = b is not what is evaluated first.
adoesn't see the inner assignment toa.