blocks|key|1129935|text|我不确定这背后是否有什么意义。这只是语言的语法。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1129936|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|$]]

I'm not sure if there is any meaning behind this. It is just the syntax of the language.

blocks|key|1345230|text|我假设这只是C%2B%2B语法的一部分。我认为对于给定的特定二进制格式，编译器如何实际实现这一点没有任何限制。对于早期的C%2B%2B编译器，您的假设可能是正确的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1345231|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|$]]

I would assume that this is just part of the C++ grammar. I don't think there are any restrictions to how the compilers actually implement this for a given specific binary format. You're assumption probably was right for the early day C++ compilers.

blocks|key|1129985|text|C%2B%2B必须有一种方法来区分纯虚函数和普通虚函数的声明。他们选择使用=+0语法。他们可以通过添加一个纯关键字轻松地完成同样的操作。但C%2B%2B非常不愿意添加新的关键字，而更喜欢使用其他机制来引入特性。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1129986|entityMap^0|X|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|$]]

C++ must have a way to distinguish a pure virtual function from a declaration of a normal virtual function. They chose to use the <code>= 0</code> syntax. They could just have easily done the same by adding a pure keyword. But C++ is pretty loath to add new keywords and prefers to use other mechanisms to introduce features.

blocks|key|1130031|text|在这种情况下，没有任何东西是“初始化”或“分配”为零的。=+0只是一个由=和0标记组成的语法结构，这与初始化或赋值完全没有关系。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1130032|它与"vtable“中的任何实际值都没有关系。C%2B%2B语言没有"vtable“之类的概念。各种“vtable”只不过是具体实现的细节而已。|1130033|entityMap^0|S|3|10|1|12|1|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Q|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|R|8|@]|D|@]|E|$]]]|I|$]]

Nothing is "initilaized" or "assigned" zero in this case. <code>= 0</code> in just a syntactical construct consisting of <code>=</code> and <code>0</code> tokens, which has absolutely no relation to either initialization or assignment.

It has no relation to any actual value in "vtable". The C++ language has no notion of "vtable" or anythng like that. Various "vtables" are nothing more than just details of specific implementations.

blocks|key|1345337|text|我记得我读到过，有趣语法的理由是它比引入另一个可以做同样事情的关键字更容易(就标准接受度而言)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1345338|我相信在Bjarne+Stroustrup的“C%2B%2B的设计和演变”中提到了这一点。|1345339|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|$]]

I remember reading that the justification for the funny syntax was that it was easier (in terms of standards-acceptance) than introducing another keyword that would do the same thing.

I believe this was mentioned in The Design and Evolution of C++ by Bjarne Stroustrup.

blocks|key|1345354|text|您还可以将vtable条目初始化为指向实际函数。“|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1345355|+virtual+void+fun()
+{
+++++//dostuff()
+}|code-block|syntax|javascript|1345356|vtable条目可以定义为无处指向(0)，也可以定义为指向函数，这似乎很直观。让你为它指定自己的值可能会导致它指向垃圾而不是函数。但这就是"=+0“被允许而"=+1”不被允许的原因。我怀疑尼尔·巴特沃斯关于为什么使用"=+0“的说法是对的。|1345357|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|$]]

Well, you can also initialize the vtable entry to point to an actual function"

<pre><code> virtual void fun()
 {
 //dostuff()
 }
</code></pre>

Seems intuitive that the vtable entry can either be defined to point nowhere (0) or to a function. Letting you specify your own value for it would probably result in it pointing to garbage instead of to a function. But that is why "= 0" is allowed and "= 1" is not. I suspect Neil Butterworth is right about why "= 0" is used at all

blocks|key|1345409|text|C%2B%2B标准的9.2节给出了类成员的语法。它包括这样的产品：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1345410|pure-specifier:
++++=+0|code-block|syntax|javascript|1345411|值并没有什么特别之处。"=+0“只是表示”这个函数是纯虚拟的“的语法。它与初始化或空指针或数值零无关，尽管与这些东西的相似性可能具有助记值。|1345412|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|$]]

Section 9.2 of the C++ standard gives the syntax for class members. It includes this production:

<pre><code>pure-specifier:
 = 0
</code></pre>

There is nothing special about the value. "= 0" is just the syntax for saying "this function is pure virtual." It has nothing to do with initialization or null pointers or the numeric value zero, although the similarity to those things may have mnemonic value.

blocks|key|1345448|text|C%2B%2B一直避免引入新的关键字，因为新的保留字打破了使用这些字作为标识符的旧程序。它通常被视为该语言的优势之一，它尽可能地尊重旧代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1345449|可能确实选择了=+0语法，因为它类似于将vtable条目设置为0，但这纯粹是象征性的。(大多数编译器将这样的vtable条目分配给存根，存根在中止程序之前会发出错误。)之所以选择语法，主要是因为它以前没有用于任何用途，而且它省去了引入新关键字的麻烦。|offset|length|style|CODE|1345450|entityMap^0|0|7|3|V|1|0^^$0|@$1|2|3|4|5|6|7|J|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|K|8|@$D|L|E|M|F|G]|$D|N|E|O|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|P|8|@]|9|@]|A|$]]]|I|$]]

C++ has always shied away from introducing new keywords, since new reserved words break old programs which use these words for identifiers. It's often seen as one of the language's strengths that it respects old code as far as possible. 

The <code>= 0</code> syntax might indeed have been chosen since it resembles setting a vtable entry to <code>0</code>, but this is purely symbolic. (Most compilers assign such vtable entries to a stub which emits an error before aborting the program.) The syntax was mainly chosen because it wasn't used for anything before and it saved introducing a new keyword.

We always declare a pure virtual function as:

<pre><code>virtual void fun () = 0 ;
</code></pre>

I.e., it is always assigned to 0.

What I understand is that this is to initialize the vtable entry for this function to NULL and any other value here results in a compile time error. Is this understanding correct or not?

Why is a pure virtual function initialized by 0?

编译器

我们总是这样声明一个纯虚函数：virtual void fun () = 0 ;也就是说，它总是赋值为0。我所理解的是，这是将此函数的vtable条目初始化为NULL，而此处的任何其他值都会导致编译时错误。这种理解是正确还是错误的？

问为什么一个纯虚函数是由0初始化的？
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