Why this code gives 7 as output?

Question

int i = 2;
i = ++i + i++ ;
cout << i;

Answer 1

My guess:
i = ++i + i++; (i is 2)
i = 3 + i++; (i is 3)
i = 3 + 3 (i is 3)
i is 6, and is then incremented by i++ to 7.

But really, you won't find this kind of stuff in real life, and I'm not even sure the implementation is standardized and not compiler-dependant.

Answer 2

according to your q.
i=++i + i++;
the compiler firstly preincrements i after seeing ++i
now i becomes 3 and the line as follows:
i=3+3;
now the post increment you used makes compiler to increment the value by 1 after executing above line , hence i becomes 7 is show on screen....
this happens because the compiler reads a whole line at once.....

Answer 3

As @Zen says, using our logic (from left to right) the result is 7.
But if the compiler uses another order (from right to left) the result is 6.
i = ++i + i++ --> i = i++ + ++i
i = 2++ + ++i
i = 2 + 4 = 6

it depends on the compiler. That's why it's better not to use such these expressions.

Answer 4

in javascript output 6
i=2;
i=++i+i++;
alert(i);

Answer 5

isnt an order fixed, ie, rtl or ltr? and what is associativity of post and pre increment operators? @marcram

Answer 6

@Manul,
as I understand it, the way it works depends only to the rules of every compiler.
Normally it works like @Zen says, but there is the possibility that the software you use uses RtL order instead of LtR.
So, as a logic exercise, @Zen's answer is the good one; otherwise, in a real practical code, it's better not to use more than one increment in the same operation.