object declaration and definition in c++

Question

Possible Duplicate:
Do the parentheses after the type name make a difference with new?

I saw someone uses the constructor like this:

class Foo
{
  public: Foo();
};

int main(){
  Foo *f= new Foo;
}

what is the difference between Foo *f= new Foo; and Foo *f= new Foo(); ?

Is this really C++? Both won't compile (both returns `Foo*`). — kennytm, Dec 20 '11 at 08:36

score 2 · Answer 1 · edited Dec 20 '11 at 08:38

2

There is no difference between those two forms of initializations. Both will call the default constructor, given that the constructor is public.

edited Dec 20 '11 at 08:38

iammilind

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answered Dec 20 '11 at 08:35

Donotalo

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There is a difference for built-ins. – sbi Dec 20 '11 at 08:52
@sbi: i wasn't talking about built ins. may be i should be more elaborate. – Donotalo Dec 20 '11 at 09:00
1

@iammilind: how is having bold text significantly better? – sehe Dec 20 '11 at 09:09
@sehe, bold text will give the heart of the answer in brief. Those who are further interested can read it in detail. Nothing much :) – iammilind Dec 20 '11 at 09:15
@Donotalo: Yes, I think this would help. – sbi Dec 20 '11 at 10:43

score 1 · Answer 2 · answered Dec 20 '11 at 08:38

1

Ỳour example probably even won't compile, you need to declare a pointer

 Foo *f = new Foo;

and there is no difference in typing new Foo or new Foo() since both run the constructor with no arguments.

answered Dec 20 '11 at 08:38

Basile Starynkevitch

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score 1 · Accepted Answer · answered Dec 20 '11 at 08:43

First of all the code you give will not compile. You need to have

Foo* f = new Foo()

Notice the asterisk.

Otherwise the two calls have the same result for non-primitive types. I have worked in companies where the () syntax is enforced by the styleguide and for a good reason: for primitive types there can be a difference:

int* p = new p;
cout << *p << endl; // the value is arbitrary i.e. behavior is undefined.
int* q = new int();
cout << *q << endl; // outputs 0.

It may be obvious here but imagine that Foo is a typedef for instance. So my advice is: always use the Foo() syntax.

object declaration and definition in c++

3 Answers3