Expressive Puzzlers - Java Puzzlers: Traps, Pitfalls, and Corner Cases - page 23

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Puzzle 9: Tweedledum
Now it's your turn to write some code. On the bright side, you have to write only two lines for this
puzzle and two lines for the next. How hard could that be? Provide declarations for the variables x
and i such that this is a legal statement:
x += i;
but this is not:
x = x + i;
Solution 9: Tweedledum
Many programmers think that the first statement in this puzzle ( x += i ) is simply a shorthand for
the second ( x = x + i ). This isn't quite true. Both of these statements are assignment expressions
[JLS 15.26]. The second statement uses the simple assignment operator ( = ), whereas the first uses a
compound assignment operator . (The compound assignment operators are += , -= , *= , /= , %= , <<= ,
>>= , >>>= , &= , ^= , and |= .) The Java language specification says that the compound assignment E1
op = E2 is equivalent to the simple assignment E1 = ( T ) (( E1 ) op ( E2 )) , where T is the type of
E1 , except that E1 is evaluated only once [JLS 15.26.2].
In other words, compound assignment expressions automatically cast the result of the
computation they perform to the type of the variable on their left-hand side. If the type of the
result is identical to the type of the variable, the cast has no effect. If, however, the type of the result
is wider than that of the variable, the compound assignment operator performs a silent narrowing
primitive conversion [JLS 5.1.3]. Attempting to perform the equivalent simple assignment would
generate a compilation error, with good reason.
To make this concrete and to provide a solution to the puzzle, suppose that we precede the puzzle's
two assignment expressions with these declarations:
short x = 0;
int i = 123456;
The compound assignment compiles without error:
x += i; // Contains a hidden cast!