C++ – error: request for member ‘..’ in ‘..’ which is of non-class type

I have a class with two constructors, one that takes no arguments and one that takes one argument.

Creating objects using the constructor that takes one argument works as expected. However, if I create objects using the constructor that takes no arguments, I get an error.

For instance, if I compile this code (using g++ 4.0.1)…

class Foo
{
  public:
    Foo() {};
    Foo(int a) {};
    void bar() {};
};

int main()
{
  // this works...
  Foo foo1(1);
  foo1.bar();

  // this does not...
  Foo foo2();
  foo2.bar();

  return 0;
}

… I get the following error:

nonclass.cpp: In function ‘int main(int, const char**)’:
nonclass.cpp:17: error: request for member ‘bar’ in ‘foo2’, which is of non-class type ‘Foo ()()’

Why is this, and how do I make it work?

Best Answer

Foo foo2();

change to

Foo foo2;

You get the error because compiler thinks of

Foo foo2()

as of function declaration with name 'foo2' and the return type 'Foo'.

But in that case If we change to Foo foo2 , the compiler might show the error " call of overloaded ‘Foo()’ is ambiguous".

Related Solutions

C++ – Inheriting constructors

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.

C++ – Initializing template base-class member types in derived-class initializer lists

The Foo_T type will not be looked up in the base class when used in the derived (Bar) constructor.

Bar (const foo_arg_t bar_arg, const a_arg_t a_arg)
: Foo<T>(bar_arg)   // base-class initializer
{
    Foo_T = TypeA(a_arg);   TypeA, etc. // Won't compile, per the standard
}

This is per the C++ standard, which says unqualified names are generally non-dependent, and should be looked up when the template is fully defined.

Since a template base class definition is not known at that time (there could be fully specialised instances of the template being pulled in later in the compilation unit), unqualified names are never resolved to names in dependent base classes.

If you need a name from a base class when templates are involved, you have to either fully qualify them, or make them implicitly dependent in your derived class.

 Foo< T >::Foo_T = TypeA(a_arg);   // fully qualified will compile

or, make it dependent

 this->Foo_T = TypeA(a_arg);

Since the this makes it template dependent, resolving the type is postponed till "phase 2" of template instantiation (and then, the base class is also fully known)

Note that if you wanted to use a function from the base class, you could have also added a using declaration..

(inside Bar())

  some_foo_func(); // wouldn't work either

  using Foo<T>::some_foo_func;
  some_foo_func(); // would work however

Best Answer

Related Solutions

C++ – Inheriting constructors

C++ – Initializing template base-class member types in derived-class initializer lists

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