Operator precedence in C for the statement z=++x||++y&&++z

Question

I was studying operator precedence and I am not able to understand how the value of x became 2 and that of y and z is

Answer 1

++ has higher priority than ||, so the whole RHS of the assignment boils down to an increment of x and an evaluation to a truth value (1).

z = ++x         ||  ++y&&++z;
    truthy (1)     never executed

This is because ++x evaluates to true and the second branch is not executed. ++x is 2 which, in a boolean context, evaluates to true or 1. z takes the value of 1, giving you the observed final state.