关于 C 中运算符优先级的混淆

Question

I got bit confused by how to interpret the precedence of operators in the following snippet:

int a,b,c,d;
a=b=c=d=1;
a=++b>1 || ++c>1 && ++d>1

The values of a,b,c,d at the end of this code snippet are 1,2,1,1 respectively. I was trying to decipher what was happening here but to no avail.

I know the precedence of ++ is higher than any other operators so why b,c, and d doesn't equal 2? According to the result I received, I guess that the expression was evaluated left to right when at the first step b is incremented to 2 therefore ++b>1 is true then because there is a logical OR the answer is returned immediately. Like if it was : (++b>1) || (++c>1 && ++d>1)

Does operator precedence have any other role other than to group operands together? What does it have to do with the order of execution for example?

Answer 1

The reason is because of Short-circuit evaluation which means that the evaluation will stop as soon as one condition is evaluated true (counting from the left).

This:

a=++b>1 || ++c>1 && ++d>1

is therefore similar to this:

if(++b > 1) {
    a = true;
} else if(++c > 1) {
    if(++d > 1) {
        a = true;
    }
}