CODEWITHSUNDEEP
javaarrays
uml
uml

Reputation: 1189

Accessing an element in an array

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.

Upvotes: 0

Views: 127

Answers (2)

arshajii
arshajii

Reputation: 129497

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

Upvotes: 3

GolezTrol
GolezTrol

Reputation: 116100

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.

Upvotes: 4

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