Why does calling a functor with an undeclared variable work?

Question

class foo {
    public:
    bool operator () (int & i) {
        return true;
    }
};

int main() {
    foo(WhyDoesThisCompile);
    return 0;
}

When passing WhyDoesThisCompile (without spaces) to the functor, the program compiles.

Why is this? I tested it on clang 4.0.0.

@TheGreatDuck there is no foo function... foo is a class. please read accepted answer — DanielCollier, Apr 15 '17 at 21:00
@TheGreatDuck no. there is no other function called foo. this code compiles as is. see https://godbolt.org/g/4tZAS0 for example — DanielCollier, Apr 15 '17 at 21:44
@TheGreatDuck dude, it is not a defaut constructor call. did you even read the answer? "You are declaring a foo, called WhyDoesThisCompile. Yes, despite the parentheses" see my link to compiler explorer to prove that it compiles. — DanielCollier, Apr 15 '17 at 21:49
@TheGreatDuck why would a constructor accept a "WhyDoesThisCompile" when it doesnt exist."Foo(WhyDoesThisCompile)" == "Foo WhyDoesThisCompile;" — DanielCollier, Apr 15 '17 at 21:56
@TheGreatDuck i still don't think you understand. this should clear things up https://godbolt.org/g/ZbfqrB — DanielCollier, Apr 15 '17 at 22:14
@TheGreatDuck your comments make no sense. Default constructors don't accept "a type of undeclared variable" (whatever that is even supposed to mean) — M.M, Apr 16 '17 at 03:09
@TheGreatDuck the lack of understanding isn't that what you're saying isn't clear, it's that it's **clearly 100% wrong** and you keep repeating it in the face of evidence, and it's hard to understand **why** you keep insisting on adding incorrect nonsense to a question that's already been adequately answered. — hobbs, Apr 16 '17 at 05:56
@TheGreatDuck: No, it is the syntax of a declaration, as has now been explained and demonstrated multiple times. There is no function call here whatsoever. And, no, there is no such thing as type "undefined" in C++. Where did you hear that there is? — Lightness Races in Orbit, Apr 16 '17 at 11:29
@TheGreatDuck there is no such type as "undefined" in C++ ; and the syntax `a(b)` has several possible meanings depending what `a` and `b` are. Also there is no built-in function called `foo`. You must just be trolling at this point — M.M, Apr 16 '17 at 12:01
@TheGreatDuck: Which is (a) completely irrelevant, and (b) not what you were claiming — Lightness Races in Orbit, Apr 17 '17 at 01:43
@TheGreatDuck: No, it isn't. I shall now quote you: _"Where is this 'foo' function that you are calling?" " You miss my point. Somewhere in the code, you have made a function call to a function called foo" "Dude... you're either invoking the default constructor or something within the C++ language. That is definitely a function call." "There is a function somewhere. It is probable just an implicitly defined function." "I'm saying it is a call to the default constructor." "it's the syntax of a function call. Clearly c++ is intending it to mean something in terms of function calling."_ — Lightness Races in Orbit, Apr 17 '17 at 13:18
@TheGreatDuck: _(cont.)_ _"There is such a thing as type 'undefined' in c++" "How could there not be an undefined type? How else does c++ recognize when something is undefined..."_ These are your claims, and every single one of them is wrong. — Lightness Races in Orbit, Apr 17 '17 at 13:18
@TheGreatDuck: The... "declaration operator"? What are you prattling on about now? — Lightness Races in Orbit, Apr 18 '17 at 01:26
Then I guess we have nothing further to say. I shall simply refer to you all the above responses to your nonsensical claims! Good night — Lightness Races in Orbit, Apr 18 '17 at 01:27

score 30 · Accepted Answer · answered Apr 15 '17 at 20:18

30

You are not invoking the functor.

You are declaring a foo, called WhyDoesThisCompile.

Yes, despite the parentheses.

I guess you meant this:

   foo()(WhyDoesThisCompile);
// ^^^^^
// temp ^^^^^^^^^^^^^^^^^^^^
//  of   invocation of op()
// type
// `foo`

… which doesn't.

answered Apr 15 '17 at 20:18

Lightness Races in Orbit

378,754
76
643
1,055

1

so this is the same as, code -> "foo WhyDoesThisCompile;" ? if so, i find that really odd. – DanielCollier Apr 15 '17 at 20:24
8

Insanely confusing and counter-intuitive. Didn't believe it was legal C++. – DeiDei Apr 15 '17 at 20:24
@BoundaryImposition I ran into this syntax accedentally i was confused that it was still compiling, thanks for the answer! – DanielCollier Apr 15 '17 at 20:32
2

@DanielCollier: No problem - enjoy the bank hol! – Lightness Races in Orbit Apr 15 '17 at 20:33
Any idea which part of the standard grammar makes this possible? I can't seem to locate it. – DeiDei Apr 15 '17 at 20:41
9

@DanielCollier, The syntax comes from C, but in general, you can put extra parentheses in declarations. You sometimes need them, too, if not using type aliases or other tricks: `int(&arr)[2];` or `void(*f)(int);` – chris Apr 15 '17 at 20:41
10

Every few weeks, we get one of those ambiguity/vexing parse questions :) – T.C. Apr 15 '17 at 22:34
4

@DeiDei see [dcl.decl]/4 where it gives the grammar for declarations - a `noptr-declarator` is a `ptr-declarator`, and `( ptr-declarator )` is a `noptr-declarator`, which is saying that you can put parentheses around any declarator. For example `int *p;` and `int (((*(((((p))))))));` are the same – M.M Apr 16 '17 at 03:13
1

Just when I thought I knew everything about C++ syntax... – Alexander Revo Apr 16 '17 at 09:54

Why does calling a functor with an undeclared variable work?

1 Answers1