How are java increment statements evaluated in complex expressions

Question

What is the output of the following code:

int x = 2;  
x += x++ * x++ * x++;  
System.out.println(x);

I understand that ++variableName is pre-increment operator and the value of variableName is incremented before it is used in the expression whereas variableName++ increments its value after the expression is executed. What I want to know is - how does this logic apply here?

possible duplicate of a += a++ * a++ * a++ in Java. How does it get evaluated? — user289086, Commented Apr 26, 2014 at 20:33

dave · Accepted Answer · 2015-10-06 01:05:07Z

Its easier to see the what is going on with x = 1 instead of 2. The output for x=1 is 7.

The key to the understanding of this is in JLS 15.7.2 which states that the every operand is fully evaluated before any part of the operation is performed.

The Java programming language guarantees that every operand of an operator (except the conditional operators &&, ||, and ? :) appears to be fully evaluated before any part of the operation itself is performed.

Thus, x++ (each of 3 times, left to right with appropriate precedence which isn't an issue here) is evaluated, then the operation * is evaluated and assigned to the original value.

x = 1 + (1 * 2 * 3)

If x starts out with 2, you get:

x = 2 + (2 * 3 * 4)

Unlike in C, this is well defined in Java and will behave the same on each invocation in any runtime.

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