++k + k++ + + k

[Rodge Thomas]

From Mughal review question 3.7:
int k = 1;
++k + k++ + + k;
I think the steps are like this (remember increment postfix has precedence over prefix):
1. k++ returns 1 to use, then increments k, so k now is 2.
++k + 1 + + k; // k is 2
2. ++k increments k and uses that value (++2) returning 3
3 + 1 + + k; //k is 3
3. now, go left to right:
so, 3 + 1 + 3 = 7
BUT, according to the book,
the expression is "parsed as
( (++k) + (k++) ) + (+k) which yields
2 + 2 + 3 = 7
I know the result is identical, but I am trying to understand operator precedence/evaluation.
Thank you for any replies.

[Valentin Crettaz]

The book is correct, first ++k is evaluated, then k++ because they are part of an normal addition which is always evaluated from left to right.
1st element: ++k
2nd element: k++
3rd element: +k
Then, the 1st is added to the 2nd element and the result of that addition is added to the 3rd element.

[jcp0.001]

test with following fragment code.
int k=1;
int k1,k2,k3,k4;
k4=(k1=++k)+(k2=k++)+(k3=+k);
result is k1=2;k2=2;k3=3!!!

[Valentin Crettaz]

jcp0.001,
Welcome to Javaranch
We'd like you to read the Javaranch Naming Policy and change your publicly displayed name (change it here) to comply with our unique rule. Thank you.

[Paul Villangca]

I've read a nice post/article by maha anna about these kinds of problems, but I can't seem to find the link. Basically, keep track of k's value and just work your way from left to right:
- when you come across a prefix operator, add 1 to k's current value, then use this value.
- when you come across a postfix operator, use k's current value in this instance, then increment the current value by 1.
Ok, so that wasn't a very good explanation... Let's use your example:

 
k = 1;
++k  +  k++  + (+k);
(2) ->  (3) ->  (3)    current value of k
 2       2       3     value used in equation
 

So using the returned values, 2 + 2 + 3 = 7.
Hope this makes it clear, if not, I hope someone can find the original post/article.

[Valentin Crettaz]

I've read a nice post/article by maha anna about these kinds of problems, but I can't seem to find the link.
Here it is:
https://coderanch.com/t/190825/java-programmer-SCJP/certification/Array

[Barkat Mardhani]

Hi Val:
Quote from you:

The book is correct, first ++k is evaluated, then k++ because they are part of an normal addition which is always evaluated from left to right.
1st element: ++k
2nd element: k++
3rd element: +k

I understand ++k and k++ but how does +k work?
Thanks
Barkat

[Valentin Crettaz]

I understand ++k and k++ but how does +k work?
the + in +k is an unary plus which has no practical effect on k's value. +k or k yields exactly the same value. For instance, if k is 4, +k is also 4.
This is not true of the unary -, though, since if k is -4, then -k is 4 !!
For more info:JLS 15.15.3 Unary Plus Operator +

[Paul Villangca]

Also, if the variable is of a type less than int (byte, short, char), the unary operator + promotes it to an int.

[Barry Gaunt]

I played around with some complex expressions, including this one, here:Marcus Green's JCHQ SCJP Discussions. Your expression is at the botton at the current time of writing this.
[ August 19, 2002: Message edited by: Barry Gaunt ]

[Anthony Villanueva]

Of course, anybody who actually writes code like ought to be quartered and hung.

[Barry Gaunt]

Actually it does hurt quite a bit.
I remember the 'C-babes' used to have very popular competitions writing code like this.
[ August 19, 2002: Message edited by: Barry Gaunt ]