Why compiler compiles from right side?

Question

My teacher, in my school ,told me that the compiler compiles a cout statement from right
eg:- //some code
B=67;
cout<<"@"<<--B;
cout<<"@"<<char(B+1)<<++B;

//output
@66@D67

in this case the program first decrements the value of B by 1 then in the next line it starts from right increments the value prints the value of char(B+1) which becomes D otherwise it should be C...

can anyone explain me this why compiler compiles from right side..?
https://code.sololearn.com/c8GjK6moEOys/?ref=ap

Answer 1

Using the ++ or -- operators in a cout statement is very unpredictable.

Another example. This was discussed in comments.

https://code.sololearn.com/cBg3yJKuyCM9/?ref=app

Answer 2

yeah in CPP droid it shows me C instead of D