blocks|key|4317706|text|不是的。将方法虚拟化是有成本的，而且C%2B%2B的理念是不让您为您没有明确声明要使用的东西付费。如果虚拟析构函数是自动生成的，那么您就会自动付出代价。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4317707|为什么不直接定义一个空的虚拟析构函数呢？|4317708|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|$]]

No. There is a cost associated with making a method virtual, and C++ has a philosophy of not making you pay for things that you don't explicitly state that you want to use. If a virtual destructor would have been generated automatically, you would have been paying the price automatically.

Why not just define an empty virtual destructor?

blocks|key|1415281|text|在C%2B%2B+11中，您可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1415282|class+MyClass
{
++//+create+a+virtual,+default+destructor
++virtual+~MyClass()+=+default;
};|code-block|syntax|javascript|1415283|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|$]]

In C++ 11 you can use:

<pre><code>class MyClass
{
 // create a virtual, default destructor
 virtual ~MyClass() = default;
};
</code></pre>

blocks|key|114012|text|是的，通过继承具有虚拟析构函数的基类。在这种情况下，您已经为多态类(例如vtable)付出了代价。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|114013|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

Yes, by inheriting from a base class with a virtual destructor. In this case, you already pay the price for a polymorphic class (e.g. vtable).

blocks|key|113983|text|不，默认情况下，所有析构函数都不是虚拟的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|113984|您需要在所有基类上定义一个虚拟析构函数|113985|除此之外。|113986|引用斯科特·迈耶斯在他的书"Effective+C%2B%2B“中的话：|113987|C%2B%2B语言标准|113988|在这个主题上异常清晰。当您尝试通过基类指针删除派生类对象时，如果基类具有非虚析构函数(如EnemyTarget所做的那样)，则结果是未定义的|blockquote|113989|113990|113991|在实践中，如果您认为有人最终可能会从虚拟析构函数创建派生类，那么使用虚拟析构函数定义类通常是一个好主意。我倾向于让所有的类都有虚拟析构函数。是的，与此相关的成本是有的，但不使其更频繁地虚拟化的成本并不超过微不足道的运行时开销。|113992|我建议，只有当你完全确定你想要它时，才让它成为非虚拟的，而不是依赖于编译器强制执行的默认的非虚拟的。你可能不同意，但是(总而言之)我最近在一些遗留代码上遇到了可怕的内存泄漏，我所做的就是将std::vector添加到一个已经存在了几年的类中。结果发现它的一个基类没有定义析构函数(默认析构函数是空的，非虚的！)因为在此之前没有内存被像这样分配，所以在那之前没有内存泄漏。很多天的调查和后来浪费的时间...|113993|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|V|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|W|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|X|8|@]|9|@]|A|$]]|$1|H|3|I|5|6|7|Y|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|Z|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|N|3|-4|5|6|7|11|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|12|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|13|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|T|$]]

No, all destructor's are by default NOT virtual.

You will need to define a virtual destructor on all the base classes

In addition to that.

To quote Scott Meyers in his book "Effective C++":

<blockquote>
 The C++ language standard is
 unusually clear on this topic. When
 you try to delete a derived class
 object through a base class pointer
 and the base class has a non-virtual
 destructor (as EnemyTarget does), the
 results are undefined
</blockquote>

In practice, it's usually a good idea to define a class with a virtual destructor if you think that someone might eventually create a derived class from it. I tend to just make all classes have virtual destructor's anyway. Yes, there is a cost associated with that, but the cost of not making it virtual more often that not out weighs a measly bit of run-time overhead.

I suggest, only make it non-virtual when you're absolutely certain that you want it that way rather than the rely on the default non-virtual that the compilers enforce. You may disagree, however (in summary) I recently had a horrid memory leak on some legacy code where all I did was add a std::vector into one of the classes that had existed for several years. It turns out that one of it's base classes didn't have a destructor defined (default destructor is empty, non-virtual!) and as no memory was being allocated like this before no memory leaked until that point. Many days of investigation and time wasted later...

blocks|key|1415201|text|Uri和Michael是对的--我要补充的是，如果你需要接触两个文件来声明和定义析构函数，那么在头文件中定义一个最小的内联函数是完全正确的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1415202|class+MyClass
{
+++//+define+basic+destructor+right+here
+++virtual+~MyClass(){}

+++//+but+these+functions+can+be+defined+in+a+different+file
+++void+FuncA();
+++int+FuncB(int+etc);
}|code-block|syntax|javascript|1415203|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|$]]

Uri and Michael are right -- I'll just add that if what's bugging you is having to touch two files to declare and define the destructor, it's perfectly all right to define a minimal one inline in the header:

<pre><code>class MyClass
{
 // define basic destructor right here
 virtual ~MyClass(){}

 // but these functions can be defined in a different file
 void FuncA();
 int FuncB(int etc);
}
</code></pre>

blocks|key|4317942|text|目前，Uri是正确的。另一方面，在类中声明了一个虚方法之后，无论如何你都要为虚表的存在付出代价。事实上，如果你的类有虚方法，但没有虚析构函数，编译器就会警告你。这可能成为自动生成默认虚拟析构函数的候选函数，而不是令人讨厌的警告。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|4317943|entityMap^0|3|3|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

Currently, Uri is right. On the other hand, after you have declared a virtual method in your class, you are paying the price for the existence of the virtual table anyway. In fact, the compiler will warn you if your class has a virtual method, but no virtual destructor. This could become a candidate for automatic generation of the default virtual destructor instead of the pesky warning.

blocks|key|4317684|text|不是的。您需要将其声明为虚拟的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4317685|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

No. You need to declare it as virtual.

blocks|key|1101171|text|有人说它是never缺省为虚拟的。事实并非如此。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1101172|1101173|即使析构函数不是继承的，如果基类声明它的析构函数是虚的，派生的析构函数总是覆盖它。|blockquote|1101174|1101175|1101176|这意味着，如果基类有一个虚拟析构函数，那么不必将派生类的析构函数定义为虚的，也不必显式地编写：|1101177|///+All+of+these+are+unnecessary+and+won't+change+the+destruction+behavior
///+as+long+as+Class+inherits+from+a+base+class+with+a+virtual+destructor
virtual+~Class()+=+default;
//+or
~Class()+override+=+default;
//+or
~Class()+{}|code-block|syntax|javascript|1101178|entityMap^0|5|5|0|0|0|0|0|7|2|L|2|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|X|8|@]|D|@]|E|$]]|$1|G|3|H|5|I|7|Y|8|@]|D|@]|E|$]]|$1|J|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]|$1|K|3|-4|5|6|7|10|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|11|8|@$9|12|A|13|B|C]|$9|14|A|15|B|C]]|D|@]|E|$]]|$1|N|3|O|5|P|7|16|8|@]|D|@]|E|$Q|R]]|$1|S|3|-4|5|6|7|17|8|@]|D|@]|E|$]]]|T|$]]

Some people stated that it is never defaulted to virtual. That's not exactly true.
<blockquote>
Even though destructors are not inherited, if a base class declares its destructor virtual, the derived destructor always overrides it.
</blockquote>
This means that if the base class has a virtual destructor you don't have to define a destructor of the derived class as virtual nor explicitly write:
<pre class="lang-cpp prettyprint-override"><code>/// All of these are unnecessary and won't change the destruction behavior
/// as long as Class inherits from a base class with a virtual destructor
virtual ~Class() = default;
// or
~Class() override = default;
// or
~Class() {}
</code></pre>

Can the default destructor be generated as a virtual destructor automatically?

If I define a base class but no default destructor, is there a default virtual destructor
generated automatically?

Can the default destructor be generated as a virtual destructor automatically?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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默认析构函数可以自动生成为虚拟析构函数吗？如果我定义了一个基类，但没有默认的析构函数，是否会自动生成一个默认的虚拟析构函数？

问默认析构函数可以自动生成为虚拟析构函数吗？
EN

回答 8

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