### Assignment operators

No. 53

#### 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;

### Tip

A:

According to Figure 135, “No binary + operator yielding byte the arithmetic + operator acting any two non-long values will always return a result of type int. Thus the expression age + 2 of type int can only be assigned to a variable age of type byte when using a cast:

byte age = 80;

age = (byte)(age + 2);

On contrary the operator += will accept any right hand integer value (even of type long!) thereby cycling through the range of byte values e.g.:

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

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