C++ 继承 - 无法访问的基础? [英] C++ inheritance - inaccessible base?

问题描述

我似乎无法使用基类作为函数参数，我是否弄乱了我的继承?

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

我的主要内容如下:

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

还有以这种方式从 Foo 继承的 Bar 类:

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

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

private:
    //snip
};

这样不行吗?我似乎无法在主函数中进行该调用

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

推荐答案

你必须这样做:

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

C++中class的默认继承类型是private，所以任何public和protected成员都来自基类仅限于 private.struct 另一方面，默认继承是 public.

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.

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