DEEP trouble in post/prefix notations

[vikas singh]

Pl. help me how to solve the following codes :
Int x=1;
Ln1 :System.out.println((++x)+(x++)+(x--)+(--x));
Ln2 :System.out.println(x);
My reasoning :
I took first bracket prefix notation, & incremented
The x vaue by 1, the new value of x becomes 2.
Now I took the last bracket prefix notation decremented
The x's present value(ie 2) by 1.The new value of x
Becomes 1.than I added like :
System.out.println ((2)+(1)+(1)+(1)); // 5
. . . I know iam not calculating it in right manner.
Is there anyone who could tell me the best method to calculate
Such interesting exp.
The output for the above is :
O/p Ln1 : 8
O/p Ln2 : 1
Thax,
Vikas S.

[Kalpesh Pandya]

Evaluate your expression from left to right and you will get your answer !

[Shailendra Guggali]

hi vikas
first thing put it as int x and not Int x
coming to yr question
start x = 1
++x = 2 so x becomes 2
x++ = 2 but after this operation x becomes 3
x-- = 3 but after this operation x becomes 2
--x = 1 so x becomes 1
adds up to : 2+2+3+1 = 8
and x = 1.

[Wasim Ahmed]

Vikas,
Follow this, as someone explained it to me.
put the operator values in the bracket and operand values outside than drop the values from the bracket, try this you never go wrong.
Int x=1;
Ln1 :System.out.println((++x)+(x++)+(x--)+(--x));
(2)2 + 2(3) + 3(2) + (1)1
Now drop the values from the bracket,
2 + 2 + 3 +1 = 8
Ln2 :System.out.println(x);
the last value of i was '1'
Hope it helps...

[vikas singh]

dear Wasim,
the formula was great . . .
Thanx,
Vikas.