CODEWITHSUNDEEP
javapost-incrementoperator-precedence
Piyush Saravagi
Piyush Saravagi

Reputation: 369

How are java increment statements evaluated in complex expressions

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?

Upvotes: 4

Views: 1256

Answers (1)

user289086
user289086

Reputation:

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.

Associated ideone if anyone wants to run it for themselves: https://ideone.com/Y2qcJ6

Upvotes: 8

Related Questions