Order of Evaluation for pre and post increment operators

[Mohnish Khiani]

Whats the order of evaluation?

 int i=0;
 int x=++i + i++ + ++i;
 System.out.println(x+" "+i);

Output :
5 3

What is the right output :
1+2+2=5
(or)
1+1+3=5

[vibhor sharma]

Initially value of i=0;

now , int x=++i
i=0 initial
i=1 final

int x= ++i + i++
i=0 initial i=1 (initial)
i=1 final i=2(final)


int x= ++i + i++ + ++i
i=0 initial i=1 (initial) i=3(initial)
i=1 final i=1 (final; but i=2 will go) i=3 (final;i 's value)

1+1+3=5 (x's value)


....hope that helps

[fred rosenberger]

Mohnish Khiani wrote:

What is the right output :
1+2+2=5
(or)
1+1+3=5

Wouldn't it have been easier to run the code yourself and find out?

[Mohnish Khiani]

i have executed the code and know the output is 5,but i wanted to know how?

That means evaluation is from left to right no matter pre or post operators?

[Rob Spoor]

Correct. That code is equivalent to this:

 int i=0;   
int x;
++i; // i == 1
x = i; // the x = ++i has now been handled; x == 1, i == 1
x += i; // x == 2, i == 1
i++; // the x = ++i + i++ has now been handled; x == 2, i == 2
++i; // x == 2, i == 3
x += i; // the x = ++i + i++ + ++i has now been handled; x == 5, i == 3
System.out.println(x+" "+i);

But this also shows how dangerous it can be to use pre and post increment operators inside other statements. It's very easy to loose track of what's going on.