Question regarding prefix & postfix operators

[pulikarthi]

If we are using variable (b) without postfix/prefix operators along with postfix/prefix on the same variable (b),
how its value being determined?

//added () for clarity

int b = 1; System.out.println((++b) + (b++) + (b));//--> output is 7

b = 1; System.out.println((++b) + (b) + (b++) ); //--> output is 6

b = 1; System.out.println((b) + (++b) + (b++)); //--> output is 5

[pbrockway2]

//added () for clarity

Just my opinion, but I think even more clarity is to be achieved by putting the // at the start of the lines...

Otherwise have a read of the description of unary operators in Oracle's Tutorial. Try predicting what will be output (paying very close attention to the actual wording of the tutorial). If your prediction differs from the actual output you will have a substantial question: be sure to post the runnable code you are using, the actual output and a description of your reasoning.

[pulikarthi]

Hi...thanks...following are my understanding.

1)

int b = 1; System.out.println((++b) + (b++) + (b));//--> actual output is 7

for the above code, am expecting the output as 5

My understanding -
b++ gets evaluated first (postfix operator has more precedence) - 1
last b gets evaluated second - 1
++b gets evaluated third - 3

2)

b = 1; System.out.println((++b) + (b) + (b++) ); //--> actual output is 6

am expecting output as 5

My understanding -
b++ gets evaluated first (postfix operator has more precedence) - 1
second b gets evaluated second - 1
++b gets evaluated third - 3

Can someone please let me know why they are returning two different values (7 and 6)?.

[Junky]

My understanding -
b++ gets evaluated first (postfix operator has more precedence) - 1
last b gets evaluated second - 1
++b gets evaluated third - 3

Incorrect.

As pre and post increment have equal precedence the statement is still evaluated left to right

[sunde887]

When you use post fix you do the add and then increment, with pre fix you increment then add. The evaluation should be evaluated left to right. If it's post fix you should pretend it's b and not increment it, for prefix you increment first.

For the line

int b = 1;
System.out.println(b++ + b + ++b);

Is evaluates from left to right and produces something like this

1 + 2 + 3

it adds b, then increments it, the second ocurance of b is now two since it got incremented, in the third part it increments b to 3 first. With that in mind try your best to properly explain how your three print lines work.

[pulikarthi]

Thanks for the reply. If we are giving same precedence to post fix and pre fix, i am able to understand how this works.

I have one more thing to ask.

Based on what i read here, post fix operator has more precedence than prefix operator.

Operators (The Java™ Tutorials > Learning the Java Language > Language Basics)

" The closer to the top of the table an operator appears, the higher its precedence."

Post fix appears first box and prefix appears in the second box.

am i understanding it incorrectly?.

[ozzyman]

all that means is, if you do something like this:

int b = 10 * 4 + 10;

since you have no brackets, should the output be 10*14=140 or 40+10=50 ?

its like BODMAS