Why the output is 19?

Question

int main()
{
int i=4,x;
x=++i + ++i + ++i;
cout<<x;
return 0;
}

Answer 1

No one can tell in case of c++.
Different compilers implement multiple pre-increment and post-increment in one statement in different ways.
There is no clear rule how to evaluate multiple increment statements in a single executable line in c++.
Try 10 different compilers and you will get at least 3-5 different outputs.
Java is clear in this case and answer is always 18 for this question.

Answer 2

I think
when
++i I=5
++i I=6
then
I+I = 12

then
++i I=7
then I+I+I = 12+7 =19

is this right

Answer 3

x = ++I(i = 5) + ++i (i = 6) + ++i(i = 7)
x = 5 + 6 + 7
x = 18

Answer 4

@tofik how??

Answer 5

initially i=4, then we add ++i(5) to it 3 times, so 4 + 5*3=19

Answer 6

I don't know. Maybe x was unset? But it is overwritten anyway. So that's not it.

In Java this outputs 18:

public static void main(String[] args) {
int i=4,x;
x=++i + ++i + ++i;
System.out.println(x);
}
As it is expected for me to sum the 5+6+7.

Answer 7

yes the output is 18

Answer 8

4+5+6

Answer 9

13

Answer 10

because is grater than input

Answer 11

var x = 9 + 10 ;