Assignment operators

Figure 123. The += operator Slide presentation

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 Slide presentation

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 Slide presentation
= 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 Slide presentation
%= Assign remainder of operands to left operand
&= Assign logical “and” of operands to left operand
|= Assign logical “or” of operands to left operand

exercise 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

Consider Figure 119, “No binary + operator yielding byte ” and Compound Assignment Operators.

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

We thus link:

  • 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.

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