blocks|key|318581|text|您正在经历扭曲，以防止进一步的子类化。为什么？记录该类不可扩展的事实，并使dtor成为非虚拟的。本着c的精神，如果有人真的想忽略你想要使用的方式，为什么要阻止他们呢？(我也从来没有看到final类/方法在java中的意义)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|318582|//Note:+this+class+is+not+designed+to+be+extended.+(Hence+the+non-virtual+dtor)
struct+DontExtened
{
++DontExtened();
++/*NOT+VIRTUAL*/
++~DontExtened();
++...
};|code-block|syntax|javascript|318583|entityMap^0|2K|5|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You are going through contortions to prevent further subclassing. Why? Document the fact that the class isn't extensible and make the dtor non-virtual. In the spirit of c, if someone really wants to ignore the way you intended this to be used why stop them? (I never saw the point of <code>final</code> classes/methods in java either).

<pre><code>//Note: this class is not designed to be extended. (Hence the non-virtual dtor)
struct DontExtened
{
 DontExtened();
 /*NOT VIRTUAL*/
 ~DontExtened();
 ...
};
</code></pre>

blocks|key|320464|text|1)是一个品味问题。如果我没看错的话，你的第二个和第三个解决方案在某些情况下会将错误从链接时转移到编译时，这通常会更好。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|320465|2)需要虚拟继承来强制责任将(虚拟)基类初始化为基类ctor不再可达的最派生类。|320466|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

1) is a matter of taste. If I see it correctly, your more fancy 2nd and 3rd solutions move the error in certain circumstances from link time to compile time, which in general should be better.

2) Virtual inheritance is needed to force the responsibility to initialize the (virtual) base class to the most derived class from where the base class ctor is no longer reachable.

blocks|key|3277423|text|你不能阻止继承(在C%2B%2B11的final关键字之前)-你只能阻止继承类的实例化。换句话说，没有办法防止：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3277424|class+A+{+...+};

class+B+:+public+A+{+...+};|code-block|syntax|javascript|3277425|你能做的最好的事情就是防止B类型的对象被实例化。在这种情况下，我建议你采纳kts的建议，记录A(或其他任何东西)不是用于继承的事实，给它一个非虚的析构函数，没有其他的虚函数，然后就到此为止。|3277426|entityMap^0|F|5|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

You can't prevent inheritance (before C++11's <code>final</code> keyword) - you can only prevent instantiation of inherited classes. In other words, there is no way of preventing:

<pre><code>class A { ... };

class B : public A { ... };
</code></pre>

The best you can do is prevent objects of type B from being instantiated. That being the case, I suggest you take kts's advice and document the fact that A (or whatever) is not intended to be used for inheritance, give it a non-virtual destructor, and no other virtual functions, and leave it at that.

blocks|key|317680|text|如果可以，我会选择第一种方法(私有构造函数)。原因是，几乎任何有经验的C%2B%2B程序员都会一眼看出这一点，并能够识别出您正在尝试防止子类化。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|317681|可能还有其他更棘手的方法来防止子类化，但在这种情况下，越简单越好。|317682|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

If you can, I'd go for the first option (private constructor). The reason is that pretty much any experienced C++ programmer will see that at a glance and be able to recognize that you are trying to prevent subclassing.

There might be other more tricky methods to prevent subclassing, but in this case the simpler the better.

blocks|key|318665|text|为了回答您的问题，您不能从CBase继承，因为在虚拟继承中，派生类需要能够直接访问从其虚拟继承的类。在这种情况下，从CBase派生的类需要直接访问CSealed，但由于构造函数是私有的，因此不能直接访问。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|318666|虽然我看不出这一切的用处(例如:停止继承)，但您可以使用模板进行泛化(我不认为它可以在所有编译器上编译，但它可以在MSVC上编译)|318667|template<class+T>
class+CSealed
{
++++friend+T;++++//+Don't+do+friend+class+T+because+it+won't+compile
++++CSealed()+{}
};

class+CBase+:+private+virtual+CSealed<CBase>
{
};|code-block|syntax|javascript|318668|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

To answer your question, you can't inherit from CBase because in virtual inheritance a derived class would need to have direct access to the class from which it was inherited virtually. In this case, a class that would derive from CBase would need to have direct access to CSealed which it can't since the constructor is private.

Though I don't see the usefulness of it all (ie: stopping inheritance) you can generalize using templates (I don't think it compiles on all compilers but it does with MSVC)

<pre><code>template&lt;class T&gt;
class CSealed
{
 friend T; // Don't do friend class T because it won't compile
 CSealed() {}
};

class CBase : private virtual CSealed&lt;CBase&gt;
{
};
</code></pre>

blocks|key|320537|text|从C%2B%2B11开始，您可以将final关键字添加到类中，例如|type|unstyled|depth|inlineStyleRanges|entityRanges|data|320538|class+CBase+final
{
...|code-block|syntax|javascript|320539|我能看到的想要这样做的主要原因(以及我来寻找这个问题的原因)是将一个类标记为非子类，这样您就可以安全地使用非虚拟析构函数并完全避免使用vtable。|320540|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

As of C++11, you can add the final keyword to your class, eg

<pre><code>class CBase final
{
...
</code></pre>

The main reason I can see for wanting to do this (and the reason I came looking for this question) is to mark a class as non subclassable so you can safely use a non-virtual destructor and avoid a vtable altogether.

blocks|key|315646|text|class+myclass;

++++class+my_lock+{
++++++++friend+class+myclass;
++++private:
++++++++my_lock()+{}
++++++++my_lock(const+my_lock&)+{}
++++};

++++class+myclass+:+public+virtual+my_lock+{
++++++++//+...
++++public:
++++++++myclass();
++++++++myclass(char*);
++++++++//+...
++++};

++++myclass+m;

++++class+Der+:+public+myclass+{+};

++++Der+dd;++//+error+Der::dd()+cannot+access
++++++++++++//+my_lock::my_lock():+private++member|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|315647|我在这里找到它是为了表示信任。我在这里张贴，只是其他人可以轻松访问http://www.devx.com/tips/Tip/38482|unstyled|offset|length|315648|entityMap|0|LINK|mutability|MUTABLE|url|http://www.devx.com/tips/Tip/38482^0|0|X|Y|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|R|8|@]|9|@$G|S|H|T|1|U]]|A|$]]|$1|I|3|-4|5|F|7|V|8|@]|9|@]|A|$]]]|J|$K|$5|L|M|N|A|$O|P]]]]

<pre><code>class myclass;

 class my_lock {
 friend class myclass;
 private:
 my_lock() {}
 my_lock(const my_lock&amp;) {}
 };

 class myclass : public virtual my_lock {
 // ...
 public:
 myclass();
 myclass(char*);
 // ...
 };

 myclass m;

 class Der : public myclass { };

 Der dd; // error Der::dd() cannot access
 // my_lock::my_lock(): private member
</code></pre>

I found it here to give credit. I am posting here just other people can easily access
<a href="http://www.devx.com/tips/Tip/38482" rel="nofollow">http://www.devx.com/tips/Tip/38482</a>

blocks|key|315660|text|详细说明Francis'+answer：如果类Bottom派生自类Middle，则负责构造虚拟继承基类(Top)的是派生最多的类(Bottom)。否则，在多重继承/死亡钻石场景中(通常使用虚拟继承)，编译器将不知道两个“中间”类中的哪一个应该构造单个基类。因此，当从Middle构造Middle时，会忽略Middle的构造函数对Top的构造函数的调用|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|315661|class+Top+{
++++public:
++++++++Top()+{}
}

class+Middle:+virtual+public+Top+{
++++public:
++++++++Middle():+Top()+{}+//+Top()+is+ignored+if+Middle+constructed+through+Bottom()
}

class+Bottom:+public+Middle+{
++++public:
++++++++Bottom():+Middle(),+Top()+{}
}|code-block|syntax|javascript|315662|因此，在您的问题中的方法2)或3)中，Bottom()不能调用Top()，因为它在Middle中是私有继承的(默认情况下，就像在您的代码中一样，但显式继承是值得的)，因此在Bottom中不可见。(source)|315663|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/2184667/194825|1|http://www.learncpp.com/cpp-tutorial/118-virtual-base-classes/^0|N|6|X|6|1G|3|1T|6|3P|6|3X|6|48|6|4K|3|4|F|0|0|0|J|8|V|5|15|6|2E|6|2Q|6|1|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]|$9|17|A|18|B|C]|$9|19|A|1A|B|C]|$9|1B|A|1C|B|C]]|D|@$9|1D|A|1E|1|1F]]|E|$]]|$1|F|3|G|5|H|7|1G|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1H|8|@$9|1I|A|1J|B|C]|$9|1K|A|1L|B|C]|$9|1M|A|1N|B|C]|$9|1O|A|1P|B|C]]|D|@$9|1Q|A|1R|1|1S]]|E|$]]|$1|M|3|-4|5|6|7|1T|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]]]

To elaborate on <a href="https://stackoverflow.com/a/2184667/194825">Francis' answer</a>: if class <code>Bottom</code> derives from class <code>Middle</code>, which virtually inherits from class <code>Top</code>, it is that most derived class (<code>Bottom</code>) that is responsible for constructing the virtually inherited base class (<code>Top</code>). Otherwise, in the multiple-inheritance/diamond-of-death scenario (where virtual inheritance is classically used), the compiler wouldn't know which of the two "middle" classes should construct the single base class. The <code>Middle</code>'s constructor's call to the <code>Top</code>'s constructor is therefore ignored when <code>Middle</code> is being constructed from <code>Bottom</code>:

<pre><code>class Top {
 public:
 Top() {}
}

class Middle: virtual public Top {
 public:
 Middle(): Top() {} // Top() is ignored if Middle constructed through Bottom()
}

class Bottom: public Middle {
 public:
 Bottom(): Middle(), Top() {}
}
</code></pre>

So, in the the approach 2) or 3) in your question, <code>Bottom()</code> can't call <code>Top()</code> because it's inherited privately (by default, like in your code, but it's worth making it explicit) in <code>Middle</code> and thus is not visible in <code>Bottom</code>. (<a href="http://www.learncpp.com/cpp-tutorial/118-virtual-base-classes/" rel="nofollow noreferrer">source</a>)

blocks|key|3064670|text|从C%2B%2B11开始，有一个干净的解决方案，我很惊讶这里没有看到它。我们可以将类设为final，以防止任何进一步的继承。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3064671|class+Foo+final+{};

class+Bar:+public+Foo+{};+//+Fails+to+compile+as+Foo+is+marked+as+final|code-block|syntax|javascript|3064672|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

From C++11 onward, there is a clean solution that I am surprise not to see here. We can make the class final preventing any further inheritance.
<pre class="lang-cpp prettyprint-override"><code>class Foo final {};

class Bar: public Foo {}; // Fails to compile as Foo is marked as final
</code></pre>

Recently one of my friend asked me how to prevent class inheritance in C++. He wanted the compilation to fail.
I was thinking about it and found 3 answers. Not sure which is the best one.
1) Private Constructor(s)
<pre><code>class CBase
{

public:

 static CBase* CreateInstance() 
 { 
 CBase* b1 = new CBase();
 return b1;
 }

private:

 CBase() { }
 CBase(CBase3) { }
 CBase&amp; operator=(CBase&amp;) { }


};
</code></pre>
2) Using <code>CSealed</code> base class, private ctor &amp; virtual inheritance
<pre><code>class CSealed
{

private:

 CSealed() {
 }

 friend class CBase;
};


class CBase : virtual CSealed
{

public:

 CBase() {
 }

};
</code></pre>
3) Using a <code>CSealed</code> base class, protected ctor &amp; virtual inheritance
<pre><code>class CSealed
{

protected:

 CSealed() {
 }

};

class CBase : virtual CSealed
{

public:

 CBase() {
 }

};
</code></pre>
All the above methods make sure that <code>CBase</code> class cannot be inherited further.
My Question is:
<ol>
<li>Which is the best method ? Any other methods available ?
</li>
<li>Method 2 &amp; 3 will not work unless the <code>CSealed</code> class is inherited virutally. Why is that ? Does it have anything to do with vdisp ptr ??
</li>
</ol>
PS:
The above program was compiled in MS C++ compiler (Visual Studio).
reference : <a href="http://www.codeguru.com/forum/archive/index.php/t-321146.html" rel="nofollow noreferrer">http://www.codeguru.com/forum/archive/index.php/t-321146.html</a>

Prevent class inheritance in C++

Visual Studio

编译器

最近，我的一个朋友问我如何在C++中防止类继承。他希望编译失败。我正在考虑这个问题，找到了3个答案。不知道哪一个是最好的。1) 私有构造函数class CBase{public: static CBase* CreateInstance()  {   CBase* b1 = new CBase();  return b...

问防止C++中的类继承
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