tags:

views:

112

answers:

3

I have some legacy code that I have to wrap, and I have come across this declaration:

class Foo : Bar
{
    // ...
};

This seems to compile under GCC. I know it's bad, but I can't change it. My question is, if no inheritance access specifier is present, how does the C++ compiler handle it?

+1  A: 

It's private.

Mehrdad Afshari
+5  A: 

For classes, the default is private.

For structs, the default is public.

Justin Ardini
+5  A: 

BTW, it is not called access modifier. It is called access specifier

$11.2/2 - "In the absence of an access-specifier for a base class, public is assumed when the derived class is defined with the class-key struct and private is assumed when the class is defined with the class-key class."

In your context, 'Bar' is a private base class of 'Foo'

Chubsdad