C++ 继承 - 无法访问的基础？
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C++ inheritance - inaccessible base?
提问by bandai
I seem to be unable to use a base class as a function parameter, have I messed up my inheritance?
我似乎无法使用基类作为函数参数，我是否搞砸了我的继承？
I have the following in my main:
我的主要内容如下：
int some_ftn(Foo *f) { /* some code */ };
Bar b;
some_ftn(&b);
And the class Bar inheriting from Foo in such a way:
类 Bar 以这种方式从 Foo 继承：
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
这不应该工作吗？我似乎无法在我的主函数中进行调用
回答by Andrew Noyes
You have to do this:
你必须这样做：
class Bar : public Foo
{
// ...
}
The default inheritance type of a classin C++ is private, so any publicand protectedmembers from the base class are limited to private. structinheritance on the other hand is publicby default.
classC++中 a 的默认继承类型是private，因此基类中的任何public和protected成员都限于private。struct另一方面，继承是public默认的。
回答by Jim Buck
By default, inheritance is private. You have to explicitly use public:
默认情况下，继承是私有的。您必须明确使用public：
class Bar : public Foo
class Bar : public Foo
