default constructor == constructor without argument?

If an object is instantiated without brackets, is it the same as having empty brackets?

Code:

---

class A { ... };
A a;

---

Code:

---

class A { ... };
A a();

---

Are they the same?

The only way I can get different "behaviours" out of them is

Code:

---

class A {
        A() {}
};

int main() {
        A a();
}

---

vs

Code:

---

class A {
        A() {}
};

int main() {
        A a;
}

---

Note that the constructor is private. Both work if it's declared public.

The first one compiles fine, but the second one gives

Quote:

---

cyberfish@cyberfish-desktop:/tmp$ g++ a.cpp
a.cpp: In function ‘int main()’:
a.cpp:2: error: ‘A::A()’ is private
a.cpp:6: error: within this context

---

Anyone care to enlighten me?

Thanks

The first version doesn't declare an object of type A. It declares a function which returns an object of type A.

Try this and see:

Code:

---

class A {
    A() {}
    public:
    void foo() {}
};

int main()
{
    A a();
    a.foo();
}

---

Uh...

Nice call!

Too much Java for me recently.

Thanks for pointing it out.

What if it's allocated on the heap?

Code:

---

class A {
    public:
    A() {}
    void foo() {}
};

int main()
{
    A *a = new A();
    a->foo();
}

---

Code:

---

class A {
    public:
    A() {}
    void foo() {}
};

int main()
{
    A *a = new A;
    a->foo();
}

---

They both compile. Are they the same?

For objects it is the same.
For built-in types however, using () zero-initialises the variable.

I see. Thanks.

Using () would zero-initialize any POD-struct as well.