Loopy Puzzlers - Java Puzzlers: Traps, Pitfalls, and Corner Cases - page 70

Solution 26: In the Loop
If you don't look at the program very carefully, you might think that it prints 100 ; after all, END is
100 more than START . If you look a bit more carefully, you will see that the program doesn't use the
typical loop idiom. Most loops continue as long as the loop index is less than the end value, but this
one continues as long as the index is less than or equal to the end value. So it prints 101 , right?
Well, no. If you ran the program, you found that it prints nothing at all. Worse, it keeps running
until you kill it. It never gets a chance to print count , because it's stuck in an infinite loop.
The problem is that the loop continues as long as the loop index ( i ) is less than or equal to
Integer.MAX_VALUE , but all int variables are always less than or equal to Integer.MAX_VALUE . It
is, after all, defined to be the highest int value in existence. When i gets to Integer.MAX_VALUE
and is incremented, it silently wraps around to Integer.MIN_VALUE .
If you need a loop that iterates near the boundaries of the int values, you are better off using a long
variable as the loop index. Simply changing the type of the loop index from int to long solves the
problem, causing the program to print 101 as expected:
for ( long i = START; i <= END; i++)
More generally, the lesson here is that int s are not integers. Whenever you use an integral type,
be aware of the boundary conditions. What happens if the value underflows or overflows? Often
it is best to use a larger type. (The integral types are byte , char , short , int , and long .)
It is possible to solve this problem without resorting to a long index variable, but it's not pretty: