What is x after 'x += x--'?

Question

Whats happening x after executing the code below;

int x = 10;
x += x--;

I'm expecting result 19 (adding x to x and then decrease x by 1) but result 20. How does it works behind the curtains? Thank your for your answers.

Answer 1

The behavior in this case was undefined before C++17, see e.g., https://en.cppreference.com/w/cpp/language/eval_order#Undefined_behavior , so if your compiler does not conform to it, it is no use testing or trying to understand it: it will depend on implementation, or even version of the compiler.

If your compiler conforms to C++17, it is guaranteed that in a simple or compound assignment (= or e.g. +=, respectively) all of the side effects of right hand side will be dealt with before evaluating the left hand side.

In your case, x-- is evaluated to be 10 accompanied by setting x=9 as its side effect, then the computer will add 10 to x=9 resulting in x=19.

Thanks to Michał for his correction, which I incorporated into the answer.

Answer 2

Having an older c++ might probably not do the job as the behaviour is literally defined to be undefined, by the older standard. (Thanks to @LightnessRacesinOrbit)

If you just try out an online compiler which will have the latest version, it works just fine and the result is 19 as you expected (x+=x-- is the same as x= x+x--. This means for getting the new "x", it has to sum the old "x"+the old "x" -1. So it will do x+(x--), which is x=10+(9).

Try it out here: with:

#include <iostream>

using namespace std;

int main()
{
    int x = 10;
    x += x--;
    cout<<x<<endl;
}