How is the output of this C program 9 and not 8?

Question

#include <stdio.h>
int main(){
int x=3,y;
y=(++x)+(x++);
printf("%d",y);
return 0;
}

The reason why i think 8 was the correct answer:
The pre-increment operator ++i is evaluated first. It increments the value of i to 4 and returns the new value, which is used in the addition. So the expression now becomes 4 + (i++).

Next, the value of i is 4, so the post-increment operator i++ returns the current value of i, which is 4, and then increments i to 5. (but the current value is 4) So the expression now becomes 4 + 4.

The addition is performed, and the result is 8.

Answer 1

Unfortunately, it is not that easy. To understand the output of your program, you must understand "sequence points". In a nutshell and our purpose here, sequence points are points in the code where all side effects, such as pre- and postincrements, are resolved. In between sequence points, the designers of C compilers are free to order the underlying CPU instructions anyway they like.

In the expression (++x)+(x++) there is no sequence point. The order of effect of the increments is undetermined. And for that reason, said expression results in undefined behaviour.

Don't try to argue why a value should result and others not. It is a moot point to argue. It is undefined behaviour. Period.