Java Reference
In-Depth Information
Compound Arithmetic Assignment Operators
Each of the five basic arithmetic operators (+, -, *, /, and %) has a corresponding compound arithmetic assignment
operator. These operators can be explained better with an example. Suppose you have two variables,
num1
and
num2
.
int num1 = 100;
byte num2 = 15;
If you want to add the value of
num1
to
num2
, you would write code as
num2 = (byte)(num2 + num1);
You need to cast the result of
num2 + num1
to
byte
because the data type of the expression is
int
. The same effect
can be rewritten using the compound arithmetic operator (+=), as follows:
num2 += num1; // Adds the value of num1 to num2
A compound arithmetic assignment operator is used in the following form:
operand1 op= operand2
Here,
op
is one of the arithmetic operators (+,
-
,
*
,
/
, and
%
).
operand1
and
operand2
are of primitive numeric
data types, where
operand1
must be a variable. The above expression is equivalent to the following expression:
operand1 = (Type of operand1) (operand1 op operand2)
For example,
int i = 100;
i += 5.5; // Assigns 105 to i
is equivalent to
i = (int)(i + 5.5); // Assigns 105 to i
There are two advantages of using the compound arithmetic assignment operators.
operand1
is evaluated only once. For example, in
i += 5.5
, the variable i is evaluated only
once, whereas in
i = (int) (i + 5.5), the variable i
is evaluated twice.
•
The
operand1
before assignment. The cast may
result in a narrowing conversion or an identity conversion. In the above example, the cast is
a narrowing conversion. The expression
i + 5.5
is of the type
double
and the result of this
expression is cast to
int
. So, the result of
double 105.5
is converted to
int 105
. If you write
an expression like
i += 5
, the equivalent expression will be
i = (int)(i + 5)
. Because the
type of the expression
i + 5
is already int, the casting the result to
int
again is an identity
conversion.
•
The result is automatically cast to the type of
