Short circuit evaluation with both && || operator

Question

I know what is short circuit evaluation in C.

a && b (operand b is not checked if a = 0)

a || b (operand b is not checked i

Answer 1

The expression 5 || 2 && ++x is equivalent to 5 || (2 && ++x) due to operator precedence.

The run time evaluates the expression 5 || 2 && ++x from left to right.

As we know in OR if first condition is true it will not check the second condition.
So here 5 evaluated as true and so (2 && ++x) will not be performed.

That's why x will remain 0 here.

Answer 2

Correct. The expression is short circuited. You can test it with this.

if(5 || ++x) {
  printf("%d\n",x);
}