Java += and implicit casting

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

Question


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.

Answers

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

or

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

or

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

or

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

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