"not declared in this scope" error with templates and inheritance [duplicate]

Question

Here is code sample which reproduces my problem:

template <typename myType>
class Base {
public:
    Base() {}
    virtual ~Base() {}
protected:
    int myOption;
    virtual void set() = 0;
};

template <typename InterfaceType>
class ChildClass : public Base < std::vector<InterfaceType> >
{
public:
    ChildClass() {}
    virtual ~ChildClass() {}
 protected:
    virtual void set();
};

template <typename InterfaceType>
void ChildClass<InterfaceType>::set()
{
     myOption = 10;
}

My usage in main():

ChildClass<int> myObject;

I get the following error (gcc 4.4.3 on ubuntu):

‘myOption’ was not declared in this scope

If my ChildClass would be without new template parameter this would work fine, i.e.:

class ChildClass : public Base < std::vector<SomeConcreteType> >

Edit

I've managed to solve it, if my set method looks like:

Base<std::vector<InterfaceType> >::myOption = 10;

It works fine. Still though not sure why I need to specify all template parameters.

myOption is a dependent name, it will work with this->

iammilind
– iammilind

2011-08-16 09:48:24 +00:00
Commented Aug 16, 2011 at 9:48 — iammilind
– iammilind, Commented Aug 16, 2011 at 9:48

Yakov Galka · Accepted Answer · 2011-08-16 09:44:22Z

66

myOption is not a dependent name, i.e. it doesn't depend on the template arguments explicitly so the compiler tries to look it up early. You must make it a dependent name:

template <typename InterfaceType>
void ChildClass<InterfaceType>::set()
{
     this->myOption = 10;
}

Now it depends on the type of this and thus on the template arguments. Therefore the compiler will bind it at the time of instantiation.

This is called Two-phase name lookup.

answered Aug 16, 2011 at 9:44

Yakov Galka

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6 Comments

iammilind Over a year ago

+1, I knew it but still missed it. :)

Joey Dumont Over a year ago

Might require another question, but is there a way to avoid having to refer to every base class variable by this->myVar? It makes the code ugly.

Yakov Galka Over a year ago

@JoeyDumont: I can't think of any. Defining a function named myOption() returning a reference to this->myOption will shorten this (no pun intended), but IMO will be even more ugly. Besides, I don't think this-> is ugly at all. This is how you would write it in C, say.

Tomas Kubes Over a year ago

It works without this in Visual Studio 2008, just myOption = 10; is OK.

Yakov Galka Over a year ago

@qub1n: it is not OK. It is Visual Studio 2008 which is not OK if it compiles this code. The authority that does specify what is correct C++ is ISO WG21, and it says that this program is invalid (see other answers for quotes).

|

Lightness Races in Orbit · Accepted Answer · 2011-08-16 10:03:53Z

19

C++03 14.6.2 Dependent names

In the definition of a class template or a member of a class template, if a base class of the class template depends on a template-parameter, the base class scope is not examined during unqualified name lookup either at the point of definition of the class template or member or during an instantiation of the class template or member.

The following code should work.

template <typename InterfaceType>
void ChildClass<InterfaceType>::set()
{
   Base<std::vector<InterfaceType> >::myOption = 10;
}

edited Aug 16, 2011 at 10:03

Lightness Races in Orbit

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answered Aug 16, 2011 at 9:45

Eric Z

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2 Comments

Lightness Races in Orbit Over a year ago

(Rhetorical question) Great, but why?

Lightness Races in Orbit Over a year ago

@Eric: Better ;) A combination of your answer and ybungalobill's would be the perfect answer.

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