C++ inheritance - inaccessible base?

Question

I seem to be unable to use a base class as a function parameter, have I messed up my inheritance?

I have the following in my main:

int some_ftn(Foo *f) { /* some code */ };
Bar b;
some_ftn(&b);

And the class Bar inheriting from Foo in such a way:

class Bar : Foo
{
public:
    Bar();
    //snip

private:
    //snip
};

Should this not work? I don't seem to be able to make that call in my main function

score 333 · Accepted Answer · edited Jul 17 '15 at 15:06

333

You have to do this:

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

The default inheritance type of a class in C++ is private, so any public and protected members from the base class are limited to private. struct inheritance on the other hand is public by default.

edited Jul 17 '15 at 15:06

Barry

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answered Jan 31 '11 at 02:22

Andrew Noyes

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I have seen some recent code with `class Bar : virtual Foo`, I could not find any documentation and details about `virtual` mode of inheritance, does anybody know about it explain it ? – Mital Vora Mar 04 '21 at 10:34

score 36 · Answer 2 · answered Jan 31 '11 at 02:25

36

By default, inheritance is private. You have to explicitly use public:

class Bar : public Foo

answered Jan 31 '11 at 02:25

Jim Buck

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To expand: In a `class` , inheritance is `private` . In a `struct` , inheritance is `public` by default. – Travis Gockel Jan 31 '11 at 02:26

C++ inheritance - inaccessible base?

2 Answers2

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