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 form E1 op= E2 is equivalent to E1 = (T) ((E1) op (E2)), where T is the type of E1, except that E1 is 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

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