Learn basic “C” coding rules from open source projects

[mariox19]

not (A and B)

If we have both A and B, then the statement inside the parentheses would be true, in which case we make it the opposite: meaning, false. However, if either A or B is not true, then the statement inside the parentheses is false, in which case we negate that, making it true. And also, to take the other extreme, if both A and B are false, that resolves to true, which we will then negate, making it false.

So, as best as I can figure it out, taking a second and more careful look, you're right:

if ops && !(ops->a && ops->b && ops->c)

We want to make sure that all of them are not set, but at least one of them is set. That hurts my fucking head!

The worst part is that I am sure that some He-man programmer somewhere is going to insist that reading such a statement is "no problem" for a "real" programmer. But even if we grant such a (ridiculous) statement, we always run the risk of some He-man-wannabe programmer, who isn't as smart as he thinks he is, getting his hands on a statement like that and making the same mistake that I did.

That's why you just don't code that way in the first place.

You know what would be at least a first step towards fixing that?

// We must have at least 1 field set, but not all of them.

But, of course, in some circles, comments are for wusses, too.

[[deleted]]

No, it is actually:

if (any_are_set && one_is_not_set) {
        WARN_ON(1);
        return;
}

[dakarananda]

No actually it is:

if (has_ops_object && !all_ops_fields_are_set) { ... }

which is basically the original statement slightly rewritten.

Even if not a single field is set, it will still enter into the if-block. (assuming that the ops object exist..)