Java: += equivalence

Question

Is:

x -= y;

equivalent to:

x = x - y;

Answer 1

No, they are NOT equivalent the way you expressed them.

short x = 0, y = 0;
x -= y;    // This compiles fine!
x = x - y; // This doesn't compile!!!
              // "Type mismatch: cannot convert from int to short"

The problem with the third line is that - performs what is called "numeric promotion" ( JLS 5.6 ) of the short operands, and results in an int value, which cannot simply be assigned to a short without a cast. Compound assignment operators contain a hidden cast!

The exact equivalence is laid out in JLS 15.26.2 Compound Assignment Operators :

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.

So to clarify some of the subtleties:

• Compound assignment expression doesn't reorder the operands
  • Left hand side stays on the left, right hand side stays on the right
• Both operands are fully-parenthesized to ensure op has the lowest precedence
  • int x = 5; x *= 2 + 1; // x == 15, not 11
• There is a hidden cast
  • int i = 0; i += 3.14159; // this compiles fine!
• The left hand side is only evaluated once
  • arr[i++] += 5; // this only increments i once

Java also has *= , /= , %= , += , -= , <<= , >>= , >>>= , &= , ^= and |= . The last 3 are also defined for booleans ( JLS 15.22.2 Boolean Logical Operators ).

Related questions

• Varying behavior for possible loss of precision
• Why doesn't Java have compound assignment versions of the conditional-and and conditional-or operators? (&&=, ||=)

Answer 2

Yes, it is. This syntax is the same in most C-derived languages.

Answer 3

Not exactly. The reason it was introduced in C was to allow the programmer to do some optimizations the compiler couldn't. For example:

A[i] += 4

used to be compiled much better than

A[i] = A[i] + 4

by the compilers of the time.

And, if "x" has side effects, eg "x++" then it is wildly different.