C++ 添加虚拟消除错误：类型“基”不是派生类的直接基
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Adding virtual removes the error : type 'base' is not a direct base of derived class
提问by nitin_cherian
Conside the following sample code below:
考虑以下示例代码：
#include <iostream>
using namespace std;
class base
{
public:
base()
{
cout << "ctor in base class\n";
}
};
class derived1 : public base
{
public:
derived1()
{
cout <<"ctor in derived1 class\n";
}
};
class derived2 : public derived1
{
public:
derived2() : base()
{
cout << "ctor in derived2 class\n";
}
};
int main()
{
derived2 d2obj;
return 0;
}
This gives the error:
这给出了错误：
error: type `base' is not a direct base of `derived2'
错误：类型“base”不是“derived2”的直接基数
Why is this error occurring? If i make the base class virtual, the error is no longer there. What is the reason for this?
为什么会发生此错误？如果我将基类设为虚拟，则错误不再存在。这是什么原因？
回答by smparkes
Because baseis not a direct base of derived2. You have to give a constructor for your direct bases, derived1in this case.
因为base不是 的直接基础derived2。derived1在这种情况下，您必须为您的直接基础提供一个构造函数。
Virtual bases are the exception. They are always initialized in leaf classes, otherwise potentially you get multiple constructor calls for the same base. So if you make basevirtual, not only can you initialize it in derived2, you must.
虚拟基地是个例外。它们总是在叶类中初始化，否则可能会为同一个基类调用多个构造函数。因此，如果您进行base虚拟化，不仅可以在derived2.
The issue arises when you have a derivation graph that is not a tree. IBMhas a pretty picture. If you don't make the base (Vin the example) virtual (everywhere) you'll have multiple copies. If you do make it virtual everywhere, you'll only have one copy, but then the direct children can not run the constructor (it would run > 1 time) and the leaf class must. For more details, best to search the web.
当您有一个不是树的派生图时，就会出现问题。IBM有一个漂亮的图片。如果您不将基础（V在示例中）设为虚拟（无处不在），您将拥有多个副本。如果你在任何地方都让它成为虚拟的，你将只有一个副本，但是直接子级不能运行构造函数（它会运行 > 1 次）并且叶类必须。更多详情，最好在网上搜索。
回答by parapura rajkumar
Change
改变
class derived2 : public derived1
{
public:
derived2() : base()
{
cout << "ctor in derived2 class\n";
}
};
to
到
class derived2 : public derived1
{
public:
derived2() : derived1()
{
cout << "ctor in derived2 class\n";
}
};
This is because you trying to call base1constructor from a grand child derived2.
这是因为您试图从大子衍生 2调用base1构造 函数。
回答by danca
You have derived1 in between. You can only call immediate parents within the initializer list
你已经在两者之间派生了1。您只能在初始化列表中调用直接父母
