Explain the output pls

Question

int x=1,y=1,z=1;
cout<<(++x||++y&&++z);
cout<<x<<y<<z;

output is 1211
according to me output should be 1222

Answer 1

What I think happening here is "Short Circuit Evaluation". That means that the compiler doesnt necessarily tests all conditions, but rather stops as soon as he got what he requires (basically).
Imagine this:
In an AND-Operator, if the first condition is false, there is no need for the compiler to check the second one because first is false, so the expression cant get true anymore anyway.
Same with OR-Operator:
If a true condition is found, the overall expression must already be true.
So in your case:
The compiler recognizes the OR-Operator and evaluates that the first condition is true, therefore the overall expression becomes true without any need to check the AND-Statement since it doesnt matter at all whether it is false or true.
As a result, y and z are not incremented, and 1211 is printed.
https://code.sololearn.com/cK4fG82x6lhP/?ref=app

Answer 2

https://www.sololearn.com/discuss/988067/?ref=app

Answer 3

|| and && have the same priority. Change in your code the position of || and &&.
You will see that x and y are increased

Answer 4

tysm everyone ☺
I have understood.

Answer 5

I still not get it
first the && operator Will be tested as it has higher priority than ||operator , so there should be increment in y and z.

Answer 6

its a Boolean expression. It never finds a reason to even make it to &&. Evaluates true instantly and returns its value

Answer 7

Naitomeas answer is the reason for that behavior. Runtime Optimization, i think.

Look at my example Jamols question.

z is increased. For && both operands have to be true

Answer 8

But why it is not checking AND statement first?

Answer 9

http://en.cppreference.com/w/cpp/language/eval_order
Furthermore:
https://stackoverflow.com/questions/5473107/operator-precedence-vs-order-of-evaluation
Might be interesting too:
http://en.cppreference.com/w/cpp/language/operator_logical