C++ spooky constructor

Question

Possible Duplicate:
Why is it an error to use an empty set of brackets to call a constructor with no arguments?

Lets have this code

class Foo {
  Foo(int) { }
};

Then we have there results:

int main() {
  Foo f1 = Foo(5); // 1: OK, explicit call
  Foo f2(5); // 2: OK, implicit call
  Foo f3(); // 3: no error, "f3 is a non-class type Foo()", how so?
  Foo f4(f1); // 4: OK, implicit call to default copy constructor
  Foo f5; // 5: expected error: empty constructor missing
}

Can you explain what's happening in case 3?

Also note that example 5 is doing what expected example 3 to do and calls the default constructor. — Joe, Dec 12 '11 at 14:33
@BjörnPollex, recently, I've taken up commenting with appropriate search terms - sometimes, it's just a matter of missing terminology, and the OP can then discover for themselves.. :) anyways, I knew someone would come in quicker with a real answer.. ;) — Nim, Dec 12 '11 at 14:36

score 12 · Answer 1 · answered Dec 12 '11 at 14:32

12

The third line is parsed as declaring a function that takes no argument and returns a Foo.

answered Dec 12 '11 at 14:32

Björn Pollex

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score 8 · Accepted Answer · answered Dec 12 '11 at 14:32

8

Foo f3(); declares a function called f3, with a return type of Foo.

answered Dec 12 '11 at 14:32

Mike Seymour

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Ah, of course! I was so focused on constructors... – Jan Turoň Dec 12 '11 at 14:35

score 5 · Answer 3 · answered Dec 12 '11 at 14:33

C++ has a rule that if a statement can be interpreted as a function declaration, it is interpreted in this way.

Hence the syntax Foo f3(); actually declares a function which takes no arguments and returns Foo. Work this around by writing Foo f3;, it will call the default constructor too (if there is one, of course).

score 4 · Answer 4 · answered Dec 12 '11 at 14:33

4

f1 invokes the copy constructor after an explicit call, you were wrong on this one
f2 is an explicit constructor call // you were wrong here too
f3 declares a function
f4 is again the copy constructor, like f1 // you're right here
f5 would calls the default constructor // you're right here again

answered Dec 12 '11 at 14:33

Xeo

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`Foo f6 = 5;` would by an implicit constructor call? IIRC. – Douglas Leeder Dec 12 '11 at 14:39
@Douglas: Yes, that would be correct. – Xeo Dec 12 '11 at 14:39

score 3 · Answer 5 · answered Dec 12 '11 at 14:35

3

This isn't what you think it is:

 Foo f3();

You may think this is an explicit call of the default constructor, but it's not. It's actually a declaration of a function named f3 which takes no parameters and returns a Foo by value.

That this is parsed as a function declaration rather than a constructor call is known as the Most Vexing Parse.

answered Dec 12 '11 at 14:35

John Dibling

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No it's not. The MVP is when you provide arguments and it still looks like a function declaration. This is not the MVP it's just the programmer being silly – Lightness Races in Orbit Nov 23 '14 at 02:34

score 2 · Answer 6 · answered Dec 12 '11 at 14:33

2

You've defined a function called f3 that returns a foo in case 3. In case 5, you have no default constructor defined, so you get an error.

answered Dec 12 '11 at 14:33

MGZero

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C++ spooky constructor

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