Can anyone explain the output of this code?

Question

#include <stdio.h>
void main()
{
int i=4, x;
x= ++i + ++i + ++i;
printf("%d",x);
}

//Output: 19

Answer 1

Frank, yup, it works

Answer 2

++i increments i once though sneeze. How do you explain the double increment? Thanks.

Answer 3

The double increment

the compiler sees 3 statements
++i + ++i
A C B

statement A : ++i now i is 5
the compile goes on with
statement B : since i is already 5, ++i makes 6
but since i is the same variable also the first i is 6
statement C : 6 + 6 = 12

Answer 4

Thanks sneeze ☺

Answer 5

As far as i know, behavior of such expressions in C++ is not defined.
In this case, i guess, its 6 + 6 + 7.

Answer 6

first - i dont do c++ but in my eyes,...
is this code really works?

Answer 7

TheEpicWobbuffe, the same goes for C (undefined behavior)

Answer 8

Yes it works but the behaviour is undefined.
So it is really bad practice to code like this.
If this is a challenge question please report it.
In my opnion this is not something that should be promoted.

What happens

x = ++i //increment i twice make i= 6
x = ++i + ++i //read 6 + 6 makes 12
x = 12 + ++i// read 12 + 7 makes 19

https://code.sololearn.com/ci36tO6Sgds0

Answer 9

I don't do C++.