Java's +=, -=, *=, /= compound assignment operators
Until today I thought that for example:
i += j;
is just a shortcut for:
i = i + j;
But what if we try this:
int i = 5;
long j = 8;
Then
i = i + j; will not compile but i += j; will compile fine.
Does it mean that in fact
i += j; is a shortcut for something like this i = (type of i) (i + j)?Answer:
As always with these questions, the JLS holds the answer. In this case §15.26.2 Compound Assignment Operators. An extract:
A compound assignment expression of the formE1 op= E2is equivalent toE1 = (T) ((E1) op (E2)), whereTis the type ofE1, except thatE1is evaluated only once.
An example cited from §15.26.2
[...] the following code is correct:short x = 3; x += 4.6;and results in x having the value 7 because it is equivalent to:short x = 3; x = (short)(x + 4.6);
In other words, your assumption is correct.
http://stackoverflow.com/questions/8710619/javas-compound-assignment-operators
