Behaviour of pre/post increment operators in Multiplication scenarios
java
Solution
int x=5;
System.out.println((x++)*x); //Gives output as 30
You first take x (x = 5) as an operand. Then it's incremented to 6 which is second operand.
int x=5;
System.out.println((++x)*x); //Gives output as 36.
You first increment x by one (x = 6) and then multiply by x => 6 * 6 = 36
Problem
Possible Duplicate: Is there a difference between x++ and ++x in java? Can anyone please explain me what is happening backyard to these statements? ``` int x=5; System.out.println((x++)*x); //Gives output as 30 int x=5; System.out.println((++x)*x); //Gives output as 36. ```