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Java += and implicit casting

· Performance Zone

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This is from two popular answers to the question Java += operator


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 thisi = (type of i) (i + j)?
I've tried googling for it but couldn't find anything relevant.


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.

And an example:

For example, 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.



A good example of this casting is using *= or /=

byte b =10;
b *=5.7;System.out.println(b);// prints 57


byte b =100;
b /=2.5;System.out.println(b);// prints 40


char ch ='0';
ch *=1.1;System.out.println(ch);// prints '4'


char ch ='A';
ch *=1.5;System.out.println(ch);// prints 'a'

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Published at DZone with permission of Peter Lawrey , DZone MVB .

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