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ezra heywood
ezra heywood

Total Post:145

Posted on    May-02-2015 12:00 AM

 Java Java 

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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)?

I've tried googling for it but couldn't find anything relevant.

Mayank Tripathi
Mayank Tripathi

Total Post:397

Posted on    May-02-2015 6:11 AM

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.

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