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Java += operator

Anonymous User 1852 02-May-2015

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.

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Updated on 02-May-2015

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Anonymous User

02-May-2015

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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Java += operator

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