Operator precedence and evaluation order

Question

I can\'t understand output of this program:

#include
using namespace std;
int main()
{
    int x = 1 , y = 1, z = 1;
    cout << ( ++x || +         


        

Answer 1

&& has a higher precedence than ||, and thus it must be evaluated first.

No. Operator precedence only determines that how the operators will be bound tighter (as if by parentheses) to its arguments when parsing an expression, it doesn't influence evaluation order. In this case, it just means ++x || ++y && ++z will be parsed as (++x) || (++y && ++z), rather than (++x || ++y) && (++z).

Note that the associativity of operator|| is left-to-right, so ++x will be evaluated at first, and (++y && ++z) won't be evaluated because of short-circuit evaluation (other than overloaded operator||).