Infinite Loop Puzzles in Java

Problem 1 : Insert code in the given code segments to make the loop infinite.

class GFG
{
      public static void main(String s[]){

           /* Insert code here */

           for (int i = start; i <= start + 1; i++) {

             /* Infinite loop */
           }
      }
}

Solution:
It looks as though it should run for only two iterations, but it can be made to loop indefinitely by taking advantage of the overflow behavior.
Integer.MAX_VALUE is the maximum value that an int can store in Java. When i gets to Integer.MAX_VALUE and is incremented, it silently wraps around to Integer.MIN_VALUE. So, we can declare variable start with 1 less than maximum value.
Following is the solution:

class GFG
{
      public static void main(String s[]){

           int start = Integer.MAX_VALUE-1;
           for (int i = start; i <= start + 1; i++) {
             /* Infinite loop */
           }
      }
}

In this, start=2147483645 (Integer.MAX_VALUE-1), and the value goes like 2147483645, 2147483646, -2147483648, -2147483647…….. and so on.

Problem 2 Insert code in the given code segments to make the loop infinite.

class GFG
{
      public static void main(String s[]) {

           /* Insert code here */

           while (i <= j && j <= i && i != j) {

            /* Infinite loop */

           }
      }
}

Solution:
Until release 5.0, Java’s numerical comparison operators (=) required both of their operands to be of a primitive numeric type (byte, char, short, int, long, float, or double). In release 5.0, the specification was changed to say that the type of each operand must be convertible to a primitive numeric type. Therein lies the rub.
In release 5.0, autoboxing and auto-unboxing were added to the language. We are using this in the following :

class GFG
{
      public static void main(String s[]){
          Integer i = new Integer(0);
          Integer j = new Integer(0);
           while (i <= j && j <= i && i != j) {
            /* Infinite loop */
           }
      }
}

The first two subexpressions (i <= j and j <= i) perform unboxing conversions on i and j and compare the resulting int values numerically. Both i and j represent 0, so both of these subexpressions evaluate to true. The third subexpression (i != j) performs an identity comparison on the object references i and j. The two variables refer to distinct objects, as each was initialized to a new Integer instance. Therefore, the third subexpression also evaluates to true, and the loop spins forever.

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