288

Why does this code:

class A
{
    public: 
        explicit A(int x) {}
};

class B: public A
{
};

int main(void)
{
    B *b = new B(5);
    delete b;
}

Result in these errors:

main.cpp: In function ‘int main()’:
main.cpp:13: error: no matching function for call to ‘B::B(int)’
main.cpp:8: note: candidates are: B::B()
main.cpp:8: note:                 B::B(const B&)

Shouldn't B inherit A's constructor?

(this is using gcc)

Rohit Vipin Mathews
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Sydius
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8 Answers8

533

If your compiler supports C++11 standard, there is a constructor inheritance using using (pun intended). For more see Wikipedia C++11 article. You write:

class A
{
    public: 
        explicit A(int x) {}
};

class B: public A
{
     using A::A;
};

This is all or nothing - you cannot inherit only some constructors, if you write this, you inherit all of them. To inherit only selected ones you need to write the individual constructors manually and call the base constructor as needed from them.

Historically constructors could not be inherited in the C++03 standard. You needed to inherit them manually one by one by calling base implementation on your own.


For templated base classes, refer to this example:

using std::vector;
    
template<class T>
class my_vector : public vector<T> {
    public:
    using vector<T>::vector; ///Takes all vector's constructors
    /* */
};
Antonio
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Suma
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    This is quite evil because for over a year that has been no compiler that can actually build the above code :-) – Mikhail Jan 21 '13 at 04:51
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    @Mikhail: Both clang and g++ should now support inheriting constructors: http://clang.llvm.org/cxx_status.html http://gcc.gnu.org/projects/cxx0x.html Recommend upvoting this one as the correct answer. – Jan Korous May 05 '13 at 11:46
  • just asking, would an inherited constructor be able to access/initialize the base class's private members? Or would I have to specify them as protected? – markasoftware Jun 22 '14 at 05:13
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    @Markasoftware The code setting those members still belongs to the base class. The 'using' declaration just means that B gains a constructor which calls through to A's constructor. So your members can be private :) – entheh Oct 20 '14 at 21:18
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    MSVC++ still doesn't support inheriting constructors: https://msdn.microsoft.com/en-us/library/hh567368.aspx – Serge Rogatch Jul 06 '15 at 12:54
  • @SergeRogatch You are correct, but it seem to be coming eventually: http://blogs.msdn.com/b/vcblog/archive/2015/04/29/c-11-14-17-features-in-vs-2015-rc.aspx – Suma Jul 06 '15 at 13:02
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    I am a person from the future! MSVC 2015 claims support. – imallett Jul 25 '15 at 04:34
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    Can you use this syntax to only inherit a specific constructor? – whoKnows Aug 05 '15 at 12:11
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    @whoKnows: No, this is all-or-nothing: http://en.cppreference.com/w/cpp/language/using_declaration – boycy Nov 07 '16 at 12:39
110

Constructors are not inherited. They are called implicitly or explicitly by the child constructor.

The compiler creates a default constructor (one with no arguments) and a default copy constructor (one with an argument which is a reference to the same type). But if you want a constructor that will accept an int, you have to define it explicitly.

class A
{
public: 
    explicit A(int x) {}
};

class B: public A
{
public:
    explicit B(int x) : A(x) { }
};

UPDATE: In C++11, constructors can be inherited. See Suma's answer for details.

nijoakim
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Avi
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17

This is straight from Bjarne Stroustrup's page:

If you so choose, you can still shoot yourself in the foot by inheriting constructors in a derived class in which you define new member variables needing initialization:

struct B1 {
    B1(int) { }
};

struct D1 : B1 {
    using B1::B1; // implicitly declares D1(int)
    int x;
};

void test()
{
    D1 d(6);    // Oops: d.x is not initialized
    D1 e;       // error: D1 has no default constructor
}

note that using another great C++11 feature (member initialization):

 int x = 77;

instead of

int x;

would solve the issue

Jean-François Fabre
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nenchev
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15

You have to explicitly define the constructor in B and explicitly call the constructor for the parent.

B(int x) : A(x) { }

or

B() : A(5) { }
Evan Teran
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grepsedawk
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8

How about using a template function to bind all constructors?

template <class... T> Derived(T... t) : Base(t...) {}
Pradu
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    Probably you should do it with perfect forwarding: template < typename ... Args > B( Args && ... args ) : A( std::forward< Args >( args ) ... ) {} – Maxim Ky Jul 31 '15 at 10:58
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    And you just broke `Derived`'s copy constructor. – Barry Sep 14 '16 at 00:45
  • Would Base's constructor have to be templated too? When you call Base(t...), then Base would have to be templated for whatever t is? – Zebrafish Dec 26 '16 at 20:57
2

Correct Code is

class A
{
    public: 
      explicit A(int x) {}
};

class B: public A
{
      public:

     B(int a):A(a){
          }
};

main()
{
    B *b = new B(5);
     delete b;
}

Error is b/c Class B has not parameter constructor and second it should have base class initializer to call the constructor of Base Class parameter constructor

Iqbal Haider
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2

Here is how I make the derived classes "inherit" all the parent's constructors. I find this is the most straightforward way, since it simply passes all the arguments to the constructor of the parent class.

class Derived : public Parent {
public:
  template <typename... Args>
  Derived(Args&&... args) : Parent(std::forward<Args>(args)...) 
  {

  }
};

Or if you would like to have a nice macro:

#define PARENT_CONSTRUCTOR(DERIVED, PARENT)                    \
template<typename... Args>                                     \
DERIVED(Args&&... args) : PARENT(std::forward<Args>(args)...)

class Derived : public Parent
{
public:
  PARENT_CONSTRUCTOR(Derived, Parent)
  {
  }
};
  • Don't do the first thing and hell don't do the macro. The first part defines copy and move constructor. – Bolpat Feb 24 '22 at 13:46
0

derived class inherits all the members(fields and methods) of the base class, but derived class cannot inherit the constructor of the base class because the constructors are not the members of the class. Instead of inheriting the constructors by the derived class, it only allowed to invoke the constructor of the base class

class A
{
    public: 
        explicit A(int x) {}
};

class B: public A
{
       B(int x):A(x);
};

int main(void)
{
    B *b = new B(5);
    delete b;
}