C++ Constructor not calling

Question

#include<iostream>  
using namespace std;  
class base {  
public:     
    base() {
        cout<<"Constructing base \n";
    }  
    virtual ~base() {
        cout<<"Destructing base \n";
    } 
};  
class derived: public base { 
public:    
    derived() {
        cout<<"Constructing derived \n";
    }    
    ~derived() {
        cout<<"Destructing derived \n";
    }
};   
int main(void) {
    derived d();     
    return 0; 
}

Why in this program its not calling constructor?

Can Anyone explain?
.......

You should call base() ctor in derived(). derived() : base() { cout << "Constructing derived"; } — bkausbk, Jul 23 '13 at 10:06
`derived d();` does _not_ define a variable, it _declares a function_ (taking no parameter and returning a `derived`). See [Variable Initialization](http://www.gotw.ca/gotw/001.htm) and [Variable Initialization – or Is It?](http://herbsutter.com/2013/05/09/gotw-1-solution/). @bkausbk Not needed, it's done implicitly by the compiler. — gx_, Jul 23 '13 at 10:10

score 5 · Answer 1 · edited Jul 23 '13 at 10:50

5

The problem is that you are declaring a function here:

// function d(), returns a derived object.
derived d();

What you need is

derived d; // C++03, C++11

or

derived d{}; // C++11 only

This is an "interesting" aspect of C++, where anything that can be parsed as a function declaration will be (provided it is in a context where a function can be declared).

See more on variable initialization in GoTW #1 Variable initialization - or is it?.

edited Jul 23 '13 at 10:50

Mike Seymour

249,747
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answered Jul 23 '13 at 10:07

juanchopanza

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does he get a warning? for this – Grijesh Chauhan Jul 23 '13 at 10:09
I've seen warnings for this in VS2012 – Neil Kirk Jul 23 '13 at 10:13
@GrijeshChauhan I don't get warnings on the platform I am on now (an ancient version of GCC), – juanchopanza Jul 23 '13 at 10:13
If you've got access to C++11 you can do `derived d{};` – Mark Ingram Jul 23 '13 at 10:15
@ juanchopanza correct it shouldn't warn t because its a correct declaration e.g `int d();` – Grijesh Chauhan Jul 23 '13 at 10:16
@MarkIngram I was just adding that :) – juanchopanza Jul 23 '13 at 10:16
1

@GrijeshChauhan There are lots and lots of compiler warnings about code that is strictly speaking correct, but probably not what the author wanted. The Most Vexing Parse is just another such case: function declarations inside another function are rarely useful, and so there's a good chance that the user meant to create an object. Clang warns about this, I think. – Sebastian Redl Jul 23 '13 at 10:26

score 4 · Answer 2 · answered Jul 23 '13 at 10:06

4

Try like this:

int main(void) {  
    derived d;     
    return 0; 
}

When you type derived d() you are declaring a function and not creating object.

answered Jul 23 '13 at 10:06

jtomaszk

9,223
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Or use `derived d{};` (c++11) – sehe Jul 23 '13 at 10:17

Shumail · Answer 3 · 2013-07-23T10:19:42.547

3

Declaration not proper for derived d();

Change it to:

int main(void) 
{
  derived d;     // note the difference     
  return 0; 
}

What you are doing -> derived d(), it declares a function. that returns instance of derived class.

See this for details: http://en.wikipedia.org/wiki/Most_vexing_parse

edited Jul 23 '13 at 10:19

answered Jul 23 '13 at 10:07

Shumail

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score 2 · Answer 4 · answered Jul 23 '13 at 10:10

2

It is because derived d() is treated as "function d, called without any arguments, returning instance of derived". Google 'most vexing parse'.

answered Jul 23 '13 at 10:10

magor

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Mateusz Andrzejewski · Answer 5 · 2013-07-23T10:21:57.493

2

int main(void)
{  
    derived d();     //here you are actually declaring a function
    return 0; 
}

You should do it like this:

derived d;

Then it will work.

edited Jul 23 '13 at 10:21

answered Jul 23 '13 at 10:12

Mateusz Andrzejewski

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score 1 · Answer 6 · answered Jul 23 '13 at 10:20

In C++, before a compiler calls a function, it must be aware of the function. Either you need to write the function in the flow before it is called or it should at least be declared before the function is called. So by calling derived d() you have declared a function called d() with return type as derived object, you never created derived object. it should have been like,

int main() {
   derived d;
   return 0

}

C++ Constructor not calling

6 Answers6