Detailed explanation
The code prints 12, and the value of x after the statement is 7. The key is to distinguish the value produced by an increment expression from the value stored in the variable. In Java, x++ is post-increment: it produces the current value of x and then increases the variable by one. By contrast, ++x is pre-increment: it increases the variable first and produces the updated value. Both forms modify x, but they contribute different values to the surrounding calculation.
Java evaluates the operands of the + operator from left to right. Starting with x equal to 5, it first evaluates the left operand, x++. This operand contributes 5 to the addition, while x becomes 6. Java then evaluates the right operand, ++x. This increases x from 6 to 7 and contributes 7. Finally, the addition combines the two operand values: 5 + 7 equals 12. That result is passed to System.out.println, which prints it followed by a line separator. The earlier value 5 is not replaced by 7 just because x changes later.
Under the hood, the important model is that evaluation preserves each operand's result while applying its side effects in the specified order. For this snippet, you can reason about the expression as separate steps: save x into a temporary left value, increment x, increment x again, save its current value into a temporary right value, and add the saved values. This is a conceptual expansion, not a requirement that the compiler create named temporary variables. Java's evaluation rules make the result well-defined rather than compiler-dependent.
Interviewers use questions like this to check evaluation order, side effects, and whether you can trace changing state precisely. Operator precedence determines how an expression is grouped; it does not mean that pre-increment must execute before a post-increment elsewhere in the expression. Although this code is valid Java, splitting updates into separate statements usually makes production code easier to read, review, and debug.
Key points to remember
- Output: 12; final value of x: 7.
- Post-increment x++ produces the old value and increments x.
- Pre-increment ++x increments x and produces the new value.
- Java evaluates the addition's left operand before its right operand.
- The addition uses the saved operand values 5 and 7.
Common mistakes candidates make
- Assuming ++x executes first because it is a prefix operator.
- Thinking post-increment waits until the entire statement finishes before changing x.
- Confusing the printed result, 12, with the final value of x, 7.
Likely follow-up questions
What would x++ + x++ print if x starts at 5?
It prints 11: the first operand contributes 5 and leaves x at 6, then the second contributes 6 and leaves x at 7. The addition therefore computes 5 + 6.
What would ++x + x++ print if x starts at 5?
It prints 12: ++x changes x to 6 and contributes 6, then x++ contributes 6 and changes x to 7. The printed result matches the original expression, but the operand values differ.
Is this expression undefined or dependent on the Java compiler?
No; Java specifies left-to-right operand evaluation and the effects of both increment operators. A conforming implementation must preserve these semantics, so this snippet prints 12.
Interview tip: State the output first, then trace each operand's contributed value and the updated value of x from left to right.