Why does having a declaration of a private base class render a type name inaccessible?

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Category:Languages

I am surprised that in the following example declaring Middle's base class private makes that name unavailable as a type in a subsequent derivation.

class Base { public:   Base(Base const& b) : i(b.i) {}    int i; };  class Middle : private Base {            //<<<<<<<<<<< public:   Middle(Base const* p) : Base(*p) {} };  class Upper : public Middle { public:   Upper(Base const* p) : Middle(p) {}    //<<<<<<<<<<< }; 

Compiling thusly with g++ (Debian 6.3.0-18+deb9u1) 6.3.0 20170516...

g++ -std=c++11 privateBase.cpp 

I get the following diagnostics:

privateBase.cpp:15:9: error: ‘class Base Base::Base’ is inaccessible within this context    Upper(Base const* p) : Middle(p) {}          ^~~~ privateBase.cpp:1:12: note: declared here  class Base {             ^ 

Clearly at the point that Base was used as Middle's base class its name was available as a type. I can understand that when Base is used to denote base class storage that should be private. But having a declaration of a private base class render a type name inaccessible seems, at the very least, unexpected.

 


This is intended; see core issue 175, which even added an example illustrating this in [class.access.spec]p5:

[ Note: In a derived class, the lookup of a base class name will find the injected-class-name instead of the name of the base class in the scope in which it was declared. The injected-class-name might be less accessible than the name of the base class in the scope in which it was declared. — end note ] [ Example:

class A { }; class B : private A { }; class C : public B {   A* p;             // error: injected-class-name A is inaccessible   ::A* q;           // OK }; 

— end example ]


This falls out of the interaction between class name injection (for rationale, see Why is there an injected class name?) and the fact that in C++ access control applies after name lookup, not before.

Comment

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