Java += and implicit casting

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'


Popular posts from this blog

Java is Very Fast, If You Don’t Create Many Objects

System wide unique nanosecond timestamps

Comparing Approaches to Durability in Low Latency Messaging Queues