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:

byte b = 0;
b = b << 1;
//    ^
// error: incompatible types: possible lossy conversion from int to byte

However, the compound assignment form is allowed:

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:

long l = 180;
l = l * 2.0f;
//    ^
// error: incompatible types: possible lossy conversion from float to long

Again, the compound assignment form is permitted:

long l = 180;
l *= 2.0f;

See Puzzle #9 in ‘Java Puzzlers: Traps, Pitfalls, and Corner Cases’ for more information.