C operator precendence: Increment and logical operators

Question

Here is my code.

#include <stdio.h>

#define PRINT3(x,y,z) printf("x=%d\ty=%d\tz=%d\n",x,y,z)

int main()
{
    int x,y,z;

    x = y = z = 1;
    ++x || ++y && ++z; PRINT3(x,y,z);

    return 0;
}

The output is,

x=2     y=1     z=1

I don't understand how this happens. What I would expect is, since ++ has higher precedence than && or ||, the prefix operations here will get evaluated first. That is, ++x, ++y and ++z will get evaluated first. So the output I expected was,

x=2     y=2     z=2

Could someone help me understand please? Thanks in advance.

I looked at this question, but I still don't understand why the compiler decides to evaluate || before evaluating all the ++s. Is this compiler dependent, or is it defined in the standard that whenever an operand of || is 1, it should be evaluated before evaluating other higher precedence operations in the other operand.

Answer 1

Higher precedence of an operator doesn't guarantee that it will be evaluated first but it guarantees that the operands will be binds (parenthesize) to it first.

++x || ++y && ++z;

will be parenthesize as

( (++x) || ( (++y) && (++z) ) );  

Since || operator guarantees the evaluation of its operand from left-to-right, (++x) will be evaluated first. Due to its short circuit behavior on becoming left operand true, right operand will not be evaluated.