### Assignment operators

Figure 123. The += operator

Increment variable by right hand value:

 int a = 4; a = a + 2; System.out.println("Value:" + a); int a = 4; a += 2; System.out.println("Value:" + a); Value:6

Figure 124. The &= operator

Logical and operation:

 boolean examSuccess = true, registered = false; examSuccess = examSuccess & registered; System.out.println( "Exam success:" + examSuccess); boolean examSuccess = true, registered = false; examSuccess &= registered; System.out.println( "Exam success:" + examSuccess); Exam success:false

Figure 125. Assignment operators #1 / 2
 = Assign right to left operand += Assign sum of operands to left operand -= Assign difference of operands to left operand *= Assign product of operands to left operand /= Assign quotient of operands to left operand

Figure 126. Assignment operators #2 / 2
 %= Assign remainder of operands to left operand &= Assign logical “and” of operands to left operand |= Assign logical “or” of operands to left operand No. 51

#### Understanding +=

Q:

Consider the following snippet:

byte age = 80;

age += 2;

This will compile and execute thereby incrementing the age variable's value by 2 as expected. The seemingly equivalent code however will not even compile:

byte age = 80;

age = age + 2; // Error: Incompatible types.
// Required: byte
// Found: int

On the other hand the += operator even accepts values exceeding a byte's upper range limit of 127:

byte age = 80;

age += 200;

So why is age += 2 perfectly legal on contrary to age = age + 2 ?

### Tip

A:

The Java® Language Specification SE 19 Edition offers a definition in its Compound Assignment Operators section:

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

We provide an example illustrating this rather condensed statement:

double d = 4.5;
byte i = 3;

i  +=  d  ;
// E1 op= E2

• i to E1

• the + operator to op

• d to E2

Our variable i aka E1 is of type int. T being E1's type the expression E1 = (T) ((E1) op (E2)) hence translates to:

i = (int)(i + d);

Back to our original example: According to Figure 119, “No binary + operator yielding byte the arithmetic + operator acting any two non-long values will always return a result of type int. The expression age + 2 (The literal 2 is of type int anyway) is thus of type int and therefore requires a cast when being assigned to our variable age of type byte:

byte age = 80;

age = (byte)(age + 2);

### Note

Since age is not being defined as final the value of the expression age + 2 is not being known at compile time. Adding final would allow for compile time evaluation and thus assignment to a second variable of type byte without requiring a cast:

final byte age = 80;         // Value cannot be altered.

byte gettingOlder = age + 2; // o.K. without cast: age + 2 can be evaluated at compile time.

On contrary regarding the age += 2 expression the equivalent is age = (byte)(age + 2):

Code Execution result
byte age = 80;
age += 200;
System.out.println("Age:" + age);
Age:24

Notice 24 being equal to 80 + 200 - 256 with 256 being a byte's number of different representable values.