Why are Static Members of Base Template Classes Inaccessible in Derived Classes?

Visibility of Base Template Class Identifiers in Derived Template Class

Consider the following code snippet:

template<typename T>
class Base
{
public:
    static const bool ZEROFILL = true;
    static const bool NO_ZEROFILL = false;
};

template<typename T>
class Derived : public Base<T>
{
public:
    Derived(bool initZero = NO_ZEROFILL);    // NO_ZEROFILL is not visible
    ~Derived();
};

When compiled with GCC g 3.4.4 (cygwin), this code compilation fails because NO_ZEROFILL is not visible to the Derived template class. This behavior can be attributed to two-phase lookup in C .

Two-Phase Lookup in C

When the compiler encounters a template declaration, it only performs a preliminary lookup for identifiers used within that template. Since the actual type for T is not determined at this stage, the compiler cannot resolve identifiers that depend on this type, such as Base::NO_ZEROFILL.

In the two-phase lookup process:

1. Preliminary lookup: Identifiers are searched for within the current scope and in previously declared namespaces.
2. Template instantiation: Once a template is instantiated with specific type parameters, a second lookup is performed to find the instantiated members and identifiers.

In this case, NO_ZEROFILL is not visible during the preliminary lookup because it depends on the unknown type T. As a result, you must explicitly specify Base::NO_ZEROFILL or this->NO_ZEROFILL in the derived class to indicate that it is a member of the base class whose type is determined at template instantiation time.

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