Loopy Puzzlers - Java Puzzlers: Traps, Pitfalls, and Corner Cases - page 80

String i = "Buy seventeen copies of Effective Java!";
The int value 0 is converted to the String value "0" and appended to the blatant plug. The
resulting string is not equal to the original as computed by the equals method, so it certainly can't
be identical, as computed by the == operator. Therefore, the boolean expression ( i != i + 0 )
evaluates to TRue and the loop never terminates.
In summary, operator overloading can be very misleading . The plus sign in the puzzle looks like
addition, but it is made to perform string concatenation by choosing the correct type for the variable
i , which is String . The puzzle is made even more misleading because the variable is named i , a
name that is usually reserved for integer variables. Good variable, method, and class names are at
least as important to program readability as good comments.
The lesson for language designers is the same as in Puzzles 11 and 13 . Operator overloading can be
confusing. Perhaps the + operator should not have been overloaded for string concatenation. It may
well be worth providing a string concatenation operator, but it doesn't have to be + .
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