is the precedence of operator ignored in 'if' conditions

Question

I'v a following code:

void main()
{
  int k, x, y, z;
  printf("\nExperiment 1:");
  x = 0, y = 0, z = 0;
  k = x++ || y++ && z++;
  printf("\nx = %d, y = %d, z = %d and k = %d\n", x, y, z, k);
  printf("\nExperiment 2:");
  x = 1, y = 0, z = 0;
  k = x++ || y++ && z++;
  printf("\nx = %d, y = %d, z = %d and k = %d\n", x, y, z, k);
}

The output:

Experiment 1: x = 1, y = 1, z = 0 and k = 0

Experiment 2: x = 2, y = 0, z = 0 and k = 1

What I've understood is: For the expression to be true, either left side or right side of '||' has to be non-zero. It starts from left. If left is non-zero, it doesn't evaluate further. If it is zero, it starts on right side. On right we have '&&'. So, we again start from left side of && and in case it is zero, the expression cannot be true and it doesn't proceed. Otherwise it evaluates the right side of '&&'

My assumption was operator && has higher precedence. So, both of its arguments should have been evaluated and then && should have been applied over it followed by evaluation of both arguments of ||.

Is compiler optimizing itself? I've used Visual Studio TC compilar with Optimization disabled.

Answer 1

I think this is covered in C11 by §6.5.14 Logical OR operator (my emphasis)

Unlike the bitwise | operator, the || operator guarantees left-to-right evaluation; if the second operand is evaluated, there is a sequence point between the evaluations of the first and second operands. If the first operand compares unequal to 0, the second operand is not evaluated.