Why don't Java's +=, -=, *=, /= compound assignment operators require casting?
Until today, I thought that for example:
i += j;
Was just a shortcut for:
i = i + j;
But 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)
?
Solution 1:
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 toE1 = (T)((E1) op (E2))
, whereT
is the type ofE1
, except thatE1
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.
Solution 2:
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'
Solution 3:
Very good question. The Java Language specification confirms your suggestion.
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);
Solution 4:
Yes,
basically when we write
i += l;
the compiler converts this to
i = (int)(i + l);
I just checked the .class
file code.
Really a good thing to know
Solution 5:
you need to cast from long
to int
explicitly
in case of i = i + l
then it will compile and give correct output. like
i = i + (int)l;
or
i = (int)((long)i + l); // this is what happens in case of += , dont need (long) casting since upper casting is done implicitly.
but in case of +=
it just works fine because the operator implicitly does the type casting from type of right variable to type of left variable so need not cast explicitly.