Logical operator execution order

Question

I have though about operator overloading and come with an interesting code:

#include <iostream>

class A {
public:
    operator bool() {
        return true;
    }

    bool operator!() {
        return false;
    }
};

int main() {
    A a;

    if (!a) {
        std::cout << "HELLO";
    } else {
        std::cout << "WORLD";
    }
    std::cout << std::endl;

    return 0;
}

What will be called first and what after? And why? Is this described anywhere in the cppreference?

P.S. For downvoters and others who think I could not execute this code by myself. I could. I did. I have changed it many times to see it's behavior. So what? This is not an explanation. I have asked for a reference which clearly states what rule this code obeys. The information how it works on my machine does not answer the question - what if this is not even portable across different environments (OSes, maybe processors, and so on)?

Nothing will be called first and after. Only `operator!` is called. — StoryTeller - Unslander Monica, Sep 26 '17 at 14:24
@mch yes I have tried it but I could not explain how it works to myself. — VP., Sep 26 '17 at 14:25
It will find the explict `! operator` and never consider conversion to `bool`. If you remove the `! operator` it will find the type conversion. — Persixty, Sep 26 '17 at 14:28
Sounds like you could use a [good C++ book](http://stackoverflow.com/questions/388242/the-definitive-c-book-guide-and-list) — NathanOliver, Sep 26 '17 at 14:28
I think the downvotes are harsh. This is a well-written question, with compilable code. Yes, you could "try it", but the OP wants an explanation. — Bathsheba, Sep 26 '17 at 14:34
@Bathsheba not a down voter, but the try it by adding a breakpoint / print statement should reveal the answer to the first part of the question, making it "why is it the case" rather than "what is the answer and why" - it shows a lack of research — UKMonkey, Sep 26 '17 at 14:41
@UKMonkey - Maybe. But the rules of C++ can be so convoluted at times, I can see why one would ask. I mean, the OP places a breakpoint. They the see the operator gets called. Cool. Now they delete the operator... and it still works. I'd ask for a clarification in the OP's place. — StoryTeller - Unslander Monica, Sep 26 '17 at 14:44
Facetiously, *every* question can be closed due to lack of research. — Bathsheba, Sep 26 '17 at 14:45
@StoryTeller "the rules of C++ can be so convoluted at times" - at times? ;) — UKMonkey, Sep 26 '17 at 14:46
@Bathsheba - Yeah. Let's just have everybody read ["Teach yourself C++ in 21 days"](http://abstrusegoose.com/strips/ars_longa_vita_brevis.png). ;) — StoryTeller - Unslander Monica, Sep 26 '17 at 14:47
@UKMonkey - unsigned integer addition is pretty straight forward :) — StoryTeller - Unslander Monica, Sep 26 '17 at 14:47
@StoryTeller until it overflows, then people have to try to work out if it's UB (like the signed integer) — UKMonkey, Sep 26 '17 at 14:49
@Bathsheba like driving a car, you only start learning after the test; when you're on your own; and when going at 90mph down the motorway wondering why on earth you can't stop in time — UKMonkey, Sep 26 '17 at 14:54

score 7 · Accepted Answer · edited Sep 26 '17 at 14:37

How it works is simple. The compiler parses the source and sees if(!a). It then checks if A defines an operator!. Just so happens that it does. So that gets called.

If it had seen if(a) it would have checked that an A is convertible to something which can be used in the condition of an if. Just so happens that it is indeed convertible.

If there had been no operator!, the compiler would have instead checked that an A can be converted to something which may be logically negated. Then the conversion to bool would have been performed.

BTW, the conversion will happen even in surprising places. For instance a + 1 will compile. Not what we want, I think. It's best to allow it only where a bool is contextually expected. You do that by marking the conversion operator explicit:

explicit operator bool() {
    return true;
}

Bathsheba · Answer 2 · 2017-09-26T14:33:52.483

4

!a is no more than syntactic sugar for a.operator!(), which you have defined: that's the compiler's preferred choice.

So the conversion to bool operator is never a candidate.

You could contrive the latter by writing

if (!(bool)a) {

edited Sep 26 '17 at 14:33

answered Sep 26 '17 at 14:28

Bathsheba

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Logical operator execution order

2 Answers2