| The compound assignment `E1 op= E2` could be mistaken for being equivalent to |
| `E1 = E1 op E2`. However, this is not the case: compound assignment operators |
| automatically cast the result of the computation to the type on the left hand |
| side. So `E1 op= E2` is actually equivalent to `E1 = (T) (E1 op E2)`, where `T` |
| is the type of `E1`. |
| |
| If the type of the expression is wider than the type of the variable (i.e. the |
| variable is a byte, char, short, or float), then the compound assignment will |
| perform a narrowing primitive conversion. Attempting to perform the equivalent |
| simple assignment would generate a compilation error. |
| |
| For example, the following does not compile: |
| |
| ```java |
| byte b = 0; |
| b = b << 1; |
| // ^ |
| // error: incompatible types: possible lossy conversion from int to byte |
| ``` |
| |
| However, the compound assignment form is allowed: |
| |
| ```java |
| byte b = 0; |
| b <<= 1; |
| ``` |
| |
| Similarly, if the expression is a floating point type (float or double), and the |
| variable is an integral type (long, int, short, byte, or char), then an implicit |
| conversion will be performed. |
| |
| Example: |
| |
| ```java |
| long l = 180; |
| l = l * 2.0f; |
| // ^ |
| // error: incompatible types: possible lossy conversion from float to long |
| ``` |
| |
| Again, the compound assignment form is permitted: |
| |
| ```java |
| long l = 180; |
| l *= 2.0f; |
| ``` |
| |
| See Puzzle #9 in 'Java Puzzlers: Traps, Pitfalls, and Corner Cases' for more |
| information. |
