x = 0; if (x1.hashCode() != x2.hashCode() ) x = x + 1; if (x3.equals(x4) ) x = x + 10; if (!x5.equals(x6) ) x = x + 100; if (x7.hashCode() == x8.hashCode() ) x = x + 1000; System.out.println("x = " + x);
Option A is incorrect because if the hashCode() values are not equal, the two objects must not be equal.
Option C is incorrect because if equals() is not true there is no guarantee of any result from hashCode().
Option D is incorrect because hashCode() will often return == even if the two objects do not evaluate to equals() being true.
(2) and (3) are incorrect because the hashCode() method is very flexible in its return values, and often two dissimilar objects can return the same hash code value.
public static void main(String[] args) { Object obj = new Object() { public int hashCode() { return 42; } }; System.out.println(obj.hashCode()); }
In this case the annoynous class is extending the Object class. Within the {} you place the methods you want for that class. After this class has been declared its methods can be used by that object in the usual way e.g. objectname.annoymousClassMethod()
public class Test { public static void main (String[] args) { String foo = args[1]; String bar = args[2]; String baz = args[3]; System.out.println("baz = " + baz); /* Line 8 */ } }And the command line invocation:
> java Test red green blue
When the program entcounters line 8 above at runtime it looks for args[3] which has never been created therefore you get an
ArrayIndexOutOfBoundsException at runtime.
public class Test { public static void main (String args[]) { String str = NULL; System.out.println(str); } }
import java.util.*; class I { public static void main (String[] args) { Object i = new ArrayList().iterator(); System.out.print((i instanceof List)+","); System.out.print((i instanceof Iterator)+","); System.out.print(i instanceof ListIterator); } }
A ListIterator can be obtained by invoking the listIterator method.
(1) and (2) are incorrect because by contract hashCode() and equals() can't be overridden unless both are overridden.
Option A is wrong. HashSet makes no guarantees as to the iteration order of the set; in particular, it does not guarantee that the order will remain constant over time.
Option B is wrong. The set can be modified.
Option D is wrong. This is a Set and not a Map.
(3) is an incorrect statement and therefore a correct answer because the hashcode for a string is computed from the characters in the string.
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