Appendix A: Operator Precedence in Java

Java has well-defined rules for evaluating expressions, including operator precedence, operator associativity, and order of evalution. We describe each of these three rules.

Precedence order.

Operator precedence specifies the manner in which operands are grouped with operators. For example, 1 + 2 * 3 is treated as 1 + (2 * 3), whereas 1 * 2 + 3 is treated as (1 * 2) + 3 because the multiplication operator has a higher precedence than the addition operator. You can use parentheses to override the default operator precedence rules.


When an expression has two operators with the same precedence, the operators and operands are grouped according to their associativity. For example 72 / 2 / 3 is treated as (72 / 2) / 3 since the division operator is left-to-right associate. You can use parentheses to override the default operator associativity rules.

Most Java operators are left-to-right associative. One notable exception is the assigment operator, which is right-to-left associative. As a result, the expression x = y = z = 17 is treated as x = (y = (z = 17)), leaving all three variables with the value 17. Recall that an assignment statement evaluates to the value of its right-hand side. Some operators are not associative. For example, the expressions (x <= y <= z) and x++-- are invalid.

Precedence and associativity of Java operators.

The table below shows all Java operators from highest to lowest precedence, along with their associativity. Most programmers do not memorize them all, and even those that do still use parentheses for clarity.

Level Operator Description Associativity
16 []
array access
member access
15 ++
unary post-increment
unary post-decrement
14 +
unary plus
unary minus
unary logical NOT
unary bitwise NOT
unary pre-increment
unary pre-decrement
13 ()
object creation
12 * / % multiplicative left-to-right
11 + -
string concatenation
10 << >>
shift left-to-right
9 < <=
> >=
relational left-to-right
8 ==
equality left-to-right
7 & bitwise AND left-to-right
6 ^ bitwise XOR left-to-right
5 | bitwise OR left-to-right
4 && logical AND left-to-right
3 || logical OR left-to-right
2 ?: ternary right-to-left
1  =   +=   -=
*=   /=   %=
&=   ^=   |=
<<=  >>= >>>=
assignment right-to-left
0 ->
lambda expression arrow right-to-left

You'll find different, but equivalent, operator precedence tables on the web and in textbooks. They disagree in inconsequential ways because some operators cannot share operands, so their relative precedence order does not matter (e.g, new and !). There is no explicit operator precedence table in the Java Language Specification. Instead, the operator precedence and associativity rules are inferred via the grammar that defines the Java language.

Order of operand evaluation.

Associativity and precedence determine in which order Java groups operands and operators, but it does not determine in which order the operands are evaluated. In Java, the operands of an operator are always evaluated left-to-right. Similarly, argument lists are always evaluated left-to-right. So, for example in the expression A() + B() * C(D(), E()), the subexpressions are evaluated in the order A(), B(), D(), E(), and C(). Although, C() appears to the left of both D() and E(), we need the results of both D() and E() to evaluate C(). It is considered poor style to write code that relies upon this behavior (and different programming languages may use different rules).

Short circuiting. Java always evaluates every operand of an operator (except &&, ||, and ?:) before the operation is performed. However, for the && and || operators, Java does not evaluate the second operand unless it is necessary to resolve the result. This allows statements like if ((s != null) && (s.length() < 10)) to work reliably. Programmers rarely use the non short-circuiting versions (& and |) with boolean expressions.

Precedence order gone awry.

Sometimes the precedence order defined in a language do not conform with mathematical norms. For example, in Microsoft Excel, -a^b is interpreted as (-a)^b instead of -(a^b). So -1^2 is equal to 1 instead of -1, which is the values most mathematicians would expect. Microsoft acknowledges this quirk as a “design choice.” One wonders whether the programmer was relying on the C precedence order in which unary operators have higher precedence than binary operators. This rule agrees with mathematical conventions for all C operators, but fails with the addition of the exponentiation operator. Once the order was established in Microsoft Excel 2.0, it could not easily be changed without breaking backward compatibility.

Associativity gone awry.

Sometimes the associativty of an operator in implemented left-to-right in some programming languages but right-to-left in others. An alarming example is exponentiation. In Wolfram Alpha and Google Sheets, the exponentiation operator is right-to-left association, so 2 ^ 2 ^ 3 is treated as 2 ^ (2 ^ 3), which is 256. However, in Matlab and Excel, the exponentiation operator is left-to-right associative, so 2 ^ 2 ^ 3 is treated as as (2 ^ 2) ^ 3, which is 64. Exponentiation is not a binary operator in Java.


  1. What is the result of the following code fragment? Explain.
    System.out.println("1 + 2 = " + 1 + 2);
    System.out.println("1 + 2 = " + (1 + 2));
    Answer: 1 + 2 = 12 and 1 + 2 = 3, respectively. If either (or both) of the operands of the + operator is a string, the other is automatically cast to a string. String concatenation and addition have the same precedence and are left-to-right associative. The parentheses in the second statement ensures that the second + operator performs addition instead of string concatenation.
  2. What does the following code fragment print?
    System.out.println(1 + 2 + "abc");
    System.out.println("abc" + 1 + 2);

    Answer: 3abc and abc12, respectively. The + operator is left-to-right associative, whether it is string concatenation or addition.

  3. Add parentheses to the following expression to make it more clear to other programmers.
    year % 4 == 0 && year % 100 != 0 || year % 400 == 0
    Answer: shows a variety of equivalent expressions, including the following reasonable alternative.
    ((year % 4 == 0) && (year % 100 != 0)) || (year % 400 == 0)

  4. What is the value of the expression 1 <= 2 <= 3? Explain.

    Answer: It leads to a compile-time error. The compiler parses the expression 1 <= 2 <= 3 as (1 <= 2) <= 3. This leads to a compile-time error because you can't use the <= operator to compare a boolean to an int. So, while the relational operators have left-to-right associativity, this property is not useful.

  5. What is the result of the following code fragment? Explain.
    boolean a = false;
    boolean b = false;
    boolean c = true;
    System.out.println(a == b == c);
    Answer: It prints true. The equality operator is left-to-right associative, so a == b evaluates to true and this result is compared to c, whihc yields true.

  6. What is the value of the expression +-~17? Explain.

    Hint: The unary operators are right-to-left associative.

  7. What is the value of the expression --17? Explain.

    Answer: It leads to a compile-time error because Java parses -- as the pre-decrement operator (and not two unary minus operators).

  8. What is the value of z after executing the following code fragment? Explain.
    int x = 5;
    int y = 10;
    int z = ++x * y--;

  9. What is the value of y after executing the following code fragment? Explain.
    int x = 10;
    int y = x+++x;

  10. What is the value of x after executing the following code fragment? Explain.
    int x = 10;

    Answer: It leads to a compile-time error. Java parses it as ++(++x). The ++x term increments the variable x and evaulates to the integer 11. This leads to a compile-time error becasue you the pre-increment operator must be applied to a variable (not an integer). So, while the pre-increment operator has right-to-left associativity, this property is not useful.