# operator precedence

pradeepbill arumalla

Greenhorn

Posts: 19

posted 12 years ago

- 0

what is the answer for the _expression ?

int k=1

++k+k++ + +k

when i put this on java editor it gave 7,but i did not understand the operator precedence,so can you please explain me!!!

->my answer was 6,when i looked in the following link for operator precedence rules

http://java.sun.com/docs/books/tutorial/java/nutsandbolts/expressions.html

thank you

pradeep

int k=1

++k+k++ + +k

when i put this on java editor it gave 7,but i did not understand the operator precedence,so can you please explain me!!!

->my answer was 6,when i looked in the following link for operator precedence rules

http://java.sun.com/docs/books/tutorial/java/nutsandbolts/expressions.html

thank you

pradeep

pradeep

Anupam Sinha

Ranch Hand

Posts: 1090

posted 12 years ago

- 0

ahhhh a classic question from Khalid and Mughal. Definitely check out the links that Anupam posted.

BUT... before you get too wrapped up in it -- go study threads or inner classes instead.

BUT... before you get too wrapped up in it -- go study threads or inner classes instead.

**Operator Precedence isn't tested on the exam**(this is according to Bert Bates himself -- and he should know.- Jess

Blog:KnitClimbJava | Twitter: jsant | Ravelry: wingedsheep

Nad Shez

Greenhorn

Posts: 9

posted 12 years ago

- 0

++ or -- if placed _before_ a variable increment or decrement the value before it takes part in the "rest of the calculation". If placed _after_ a variable, increment or decrement after it takes part in the "rest of the calculation". What this means is that if a variable is appearing twice in a statement, the second time it appears, it would have the "updated" value. e.g.

int i = 5, j = 2, k;

k = i++ + j;

i=6, j=2, & k=7, right? Yes. since i is incremented after i++ is executed, but for understanding point of view, value of i is set to 6 not by the end of the statement, but by the end of when i++ is done .. meaning when it gets to the + operator in the middle. so

k = i++ + i;

would give us k=11 .. since i++ would evaluate to 5, but before the expression is further evaluated, i is set to 6 .. so when the right side of the + operator is evaluated & value of i is requested, its 6 .. hence 5+6 would give us 11. value of i by the end of expression would still be 6.

now to the main course

int k = 1;

++k+k++ + +k;// ++k + k++ + +k

Evaluating it from left to right step by step

++k yields 2 & sets the value of k = 2

2 + k++ + +k;

k++ yields 2 but sets the value of k = 3

2 + 2 + +k;

+k is a simple + Unary operator that has no effect in this case so it yields the current value of k i-e 3

2 + 2 + 3;

equals 7.

Hope that helps. please correct me if I'm wrong in my crude explanation.

int i = 5, j = 2, k;

k = i++ + j;

i=6, j=2, & k=7, right? Yes. since i is incremented after i++ is executed, but for understanding point of view, value of i is set to 6 not by the end of the statement, but by the end of when i++ is done .. meaning when it gets to the + operator in the middle. so

k = i++ + i;

would give us k=11 .. since i++ would evaluate to 5, but before the expression is further evaluated, i is set to 6 .. so when the right side of the + operator is evaluated & value of i is requested, its 6 .. hence 5+6 would give us 11. value of i by the end of expression would still be 6.

now to the main course

int k = 1;

++k+k++ + +k;// ++k + k++ + +k

Evaluating it from left to right step by step

++k yields 2 & sets the value of k = 2

2 + k++ + +k;

k++ yields 2 but sets the value of k = 3

2 + 2 + +k;

+k is a simple + Unary operator that has no effect in this case so it yields the current value of k i-e 3

2 + 2 + 3;

equals 7.

Hope that helps. please correct me if I'm wrong in my crude explanation.

Thomas Paul

mister krabs

Ranch Hand

Ranch Hand

Posts: 13974

posted 12 years ago

That isn't quite right.

If an increment/decrement is before the variable:

1) update the variable

2) return the updated variable

If an increment/decrement is after the variable:

1) return the variable

2) update the variable

As far as this extreme example, I have added parenthesis to show operator precendence:

(++k)+(k++) + (+k)

++k

return 2, k is now 2

2+(k++) + (+k)

k++ return 2, k is now 3

2+2 + (+k)

4 + (+3)

4 + 3

7

- 0

**++ or -- if placed _before_ a variable increment or decrement the value before it takes part in the "rest of the calculation". If placed _after_ a variable, increment or decrement after it takes part in the "rest of the calculation".**

That isn't quite right.

If an increment/decrement is before the variable:

1) update the variable

2) return the updated variable

If an increment/decrement is after the variable:

1) return the variable

2) update the variable

As far as this extreme example, I have added parenthesis to show operator precendence:

(++k)+(k++) + (+k)

++k

return 2, k is now 2

2+(k++) + (+k)

k++ return 2, k is now 3

2+2 + (+k)

4 + (+3)

4 + 3

7

Associate Instructor - Hofstra University

Amazon Top 750 reviewer - Blog - Unresolved References - Book Review Blog

pradeepbill arumalla

Greenhorn

Posts: 19

posted 12 years ago

- 0

hey thomas paul ,

why did u do the pre-increment first...according to the precedence rules u r supposed to do the post-increment first then start with the left- associative equal precedence of ++k &+k ..

paul,we are supposed to follow the rules & arrive at the answer but not from the answer to the question is what i feel.

thanks

why did u do the pre-increment first...according to the precedence rules u r supposed to do the post-increment first then start with the left- associative equal precedence of ++k &+k ..

paul,we are supposed to follow the rules & arrive at the answer but not from the answer to the question is what i feel.

thanks

pradeep

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