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int x = 3; 
System.out.println(x++ + ++x + x++); // 13

Why is the result 13?

My logic:

  1. ++x in the center gives 4
  2. 4 + 4 + 4 = 12 so the result must be 12.
Tunaki
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Adam Ary
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    Let's never write code like this. No one should ever use such a construct in production code. – duffymo Nov 03 '15 at 11:25
  • This is fun to analyse in Java as it's well-defined. Note that in C and C++, the behaviour is undefined. – Bathsheba Nov 03 '15 at 11:27
  • x++ does post increment. So in the evaluation the value is 13 – Vivek Singh Nov 03 '15 at 11:27
  • System.out.println(3 + 5 + 5); you should first learn basics about increment operator before asking such questions... – Shailesh Yadav Nov 03 '15 at 11:28
  • You even asked a similiar question beforehand where @Eran answered you how the post- and preincrement are working. What is confusing you that you still think that the first two operations are still `4+4`? – SomeJavaGuy Nov 03 '15 at 11:44
  • You asked a very similar question some hours ago, got an answer and still don't understand it? You still think Java evaluates your expression in a random/unreasonable order ... – Tom Nov 03 '15 at 12:15

3 Answers3

7

Let's take it step-by-step. First it is important to note that expressions are evaluated from left to right, so there is no undefined behaviour.

int x = 3;
int res = x++ + ++x + x++
// res = (x++) + (++x) + (x++) with x = 3
// res =   3   + (++x) + (x++) with x = 4
// res =   3   +   5   + (x++) with x = 5
// res =   3   +   5   +   5   with x = 6
// res = 13

The key part here is that:

  • x++ returns the previous x value and increments x afterwards. JLS section 15.14.2 about the "Postfix Increment Operator ++" says:

    The value of the postfix increment expression is the value of the variable before the new value is stored.

  • ++x returns the next x incremented value. JLS section 15.15.1 about the "Prefix Increment Operator ++" says:

    The value of the prefix increment expression is the value of the variable after the new value is stored.

Tunaki
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3
System.out.println(x++ + ++x + x++); // 13 

                    3     5     5
System.out.println(x++ + ++x + x++); // 13

x++ - executes then increments, so 3

++x - above 3 incremented to 4, ++x increments first then executes, so 4+1=5

x++ - executes first increment later, so x=5 is used, but after this execution x becomes 6

Ankur Singhal
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2

x++ is an expression evaluating to the value before the increment. When you see x++ you can think of it as

int result = x;
x = x + 1;
// do something with result.

++x can be thought of as

x = x + 1;
int result = x;
// do something with result.

The expression in the question is evaluated from left to right. You can think of it as

int x = 3

int result1 = x;  // 3
x = x + 1;

x = x + 1;
int result2 = x;  // 5

int result3 = x;  // 5
x = x + 1;

System.out.println(result1 + result2 + result3);
Paul Boddington
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