int x = ++n + n++; returns an unexpected value

Question

According to the concept of pre-increment and post-increment operator the output of the following code should be (8+8) = 16, but in the compiler it is evaluated to 17. Please explain with steps.

#include <iostream>

using namespace std;

int main()
{
    int n = 7;
    int x = ++n + n++;
    cout << x;

    return 0;
}

Take a look at [this question](https://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points). — hnefatl, Jun 11 '21 at 22:51

score 4 · Answer 1 · answered Jun 11 '21 at 22:47

4

following code should be (8+8) = 16

That's not correct.

Please explain with steps.

This expression:

++n

Modifies n.

This expression:

n++

Modifies n.

This expression:

++n + n++

Has two modifications on the same variable that are unsequenced in relation to each other. Hence, the behaviour of the program is undefined.

Since the bahviour is undefined, there is no output that "should" or "shouldn't" be.

answered Jun 11 '21 at 22:47

eerorika

232,697
12
197
326

Is this still UB in C++17? – Costantino Grana Jun 12 '21 at 00:42
1

@CostantinoGrana Yes, it is still UB. – eerorika Jun 12 '21 at 00:43
Thanks. [This page](https://en.cppreference.com/w/cpp/language/eval_order) has all the rules, including the new ones. `i = ++i + 2;` is instead well defined. – Costantino Grana Jun 12 '21 at 00:53

int x = ++n + n++; returns an unexpected value

1 Answers1