Templates: Why does this fail to compile?

[A Moore]

///
/// Both g++ 3.4.1 and comeau 4.3.3 report "i undefined"
///
#include <iostream>

template<typename T>
class Base
{
protected:
int i;
};


template<typename T>
class Derived : public Base< T >
{
public:
void f() {
i = 0;
std::cout << i << std::endl;
}
};


int main()
{
Derived< int > d;
d.f();
return 0;
}

 

[Alf P. Steinbach]

* A Moore:

///
/// Both g++ 3.4.1 and comeau 4.3.3 report "i undefined"
///
#include <iostream>

template<typename T>
class Base
{
protected:
int i;
};


template<typename T>
class Derived : public Base< T >
{
public:
void f() {
i = 0;
std::cout << i << std::endl;
}
};

Here 'i' could conceivably refer to a global variable.

Preferred solution: 'using Base<T>::i;'.

Alternative solution: 'this->i'.

 

[Andrey Tarasevich]

#include <iostream>

template<typename T>
class Base
{
protected:
int i;
};


template<typename T>
class Derived : public Base< T >
{
public:
void f() {
i = 0;

In this context non-qualified name 'i' is not a dependent name.
Non-dependent names are looked up immediately in accordance with usual
look-up rules, except that dependent base classes are excluded from this
process. That's why 'i' is not found and the code fails to compile. If
declare an 'i' variable at namespace scope, the code will compile, but
that 'i' will refer to the namespace variable, not to the base class member.

If you want your 'i' to refer to the 'i' in the base class, you have to
explicitly turn it into a dependent name. This can be done by either using

this->i = 0;

or by using a qualified name

 

[A Moore]

Well this->i does not seem very dependent to me...
How come?

 

[Andrey Tarasevich]

...
Well this->i does not seem very dependent to me...
How come?
...

'Base<T>' is a dependent type for obvious reasons.

'Derived' is a dependent type in accordance with 14.6.2.1/1 - it is a
compound type constructed from a dependent type ('Base<T>').

'this' is a type-dependent expression in accordance with 14.6.2.2/2 -
class type of the enclosing member function ('Derived') is dependent.

'this->i' is a type-dependent expression in accordance with 14.6.2.2/3 -
its subexpression ('this') is type-dependent.

 

[Sumit Rajan]

///
/// Both g++ 3.4.1 and comeau 4.3.3 report "i undefined"
///
#include <iostream>

template<typename T>
class Base
{
protected:
int i;
};


template<typename T>
class Derived : public Base< T >
{
public:
void f() {
i = 0;
std::cout << i << std::endl;
}
};


int main()
{
Derived< int > d;
d.f();
return 0;
}

Try the FAQ:
http://www.parashift.com/c++-faq-lite/templates.html#faq-35.12

(And the next question as well)

Regards,
Sumit.

 

[Greg Comeau]

Well this->i does not seem very dependent to me...
How come?

Because in the example given it depends upon a Derived<T>
(though the unadorned i may not).
See http://www.comeaucomputing.com/techtalk/templates/#whymembernotfound

 

[uououo]

All right on vs.net 2003.

 

[A Moore]

Thank you! This was very informative and impressive.

The issue looks very unnatural to me, but then
again I am not well versed in the standard.

 

[A Moore]

Thanks for your help!

* A Moore:



Here 'i' could conceivably refer to a global variable.

Preferred solution: 'using Base<T>::i;'.

Alternative solution: 'this->i'.

 

[A Moore]

Thanks!

Because in the example given it depends upon a Derived<T>
(though the unadorned i may not).
See http://www.comeaucomputing.com/techtalk/templates/#whymembernotfound