Answer for Voja

            
You already now that:
- Postfix: passes the current value of X to the function and then increments it.
- Prefix: increments the current value X and then passes it to the function.


But your example is very interesting. Let me explain to you what's happening in your code:

To avoid confusion, let x = 2 instead of x = 1.
So, in case x=2, your question will be why x+=x++ is 4 instead 5, right?


So, let we see: 🤓⤵️

int x=2;
x += x++;  is same as:  x = x + x++ ;

🔹 x = (x) + x++ -> the first x gets evaluated and returns 2, x = 2
🔹 x = 2 + (x)++ -> x gets evaluated and returns 2, x = 2
🔹 x = 2 + (x++) -> x++ gets evaluated and x is incremented (to 3), x = 3
🔹 x = (2 + 2) -> 2 + 2 gets evaluated and returns 4, x = 3
🔹 (x = 4) -> x is assigned to 4 (overriding the previous value of 3)
    


Now let’s get to this one: 🤓⤵️

int x = 2;
x = x++  + x;

🔹 x = (x)++ + x -> x gets evaluated and returns 2, x =2
🔹 x = (x++) + x -> x++ gets evaluated and x is incremented, x=3
🔹 x = 2 + (x) -> x gets evaluated and returns 3, x = 3
🔹 x = 2 + 3 -> 2+3 gets evaluated and returns 5, x = 3
🔹 (x=5) -> x is assigned to 5 (overriding the previous value of 3)



Now let’s see how this one works: 🤓⤵️    

int x = 2;
x = x +  ++x;

🔹 x = (x) + ++x -> x gets evaluated and returns 2, x=2
🔹 x = 2 + (++x) -> ++x gets evaluated and x is incremented, x=3
🔹 x = 2 + (x) -> x gets evaluated and returns 3, x=3
🔹 x = 2 + 3 -> 2+3 gets evaluated and returns 5, x = 3
🔹 (x=5) -> x is assigned to 5 (overriding the previous value of 3)
    
    
    
End, finally: 🤓⤵️

int x = 2;
x = ++x + x;

🔹 x = (++x) + x -> ++x gets evaluated and x is incremented, x=3
🔹 x = 3 + (x) -> x gets evaluated and returns 3, x=3
🔹 x = 3 + 3 -> 3+3 gets evaluated and returns 6, x = 3
🔹 (x=6) -> x is assigned to 6 (overriding the previous value of 3)
    
    
I hope this is clear now 😉.
for Voja: postfix/prefix increment
