blocks|key|320015|text|在C%2B%2B语言中，完全限定名仅用于引用现有的(即先前声明的)实体。它们不能用来引入新的实体。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|320016|实际上，您正在“重新打开”名称空间来声明新的实体。如果Class类后来被定义为不同名称空间的成员-它是一个完全不同的类，与您在此处声明的那个类没有任何关系。|offset|length|style|CODE|320017|一旦到了定义预声明类的阶段，就不需要再次“重新打开”名称空间。您可以在全局命名空间(或包含Namespace的任何命名空间)中将其定义为|320018|class+Namespace::Class+{
++/*+whatever+*/
};|code-block|syntax|javascript|320019|由于您引用的实体已在命名空间Namespace中声明，因此可以使用限定名称Namespace::Class。|320020|entityMap^0|0|R|5|0|19|9|0|0|E|9|11|G|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|W|8|@$D|X|E|Y|F|G]]|9|@]|A|$]]|$1|J|3|K|5|L|7|Z|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|10|8|@$D|11|E|12|F|G]|$D|13|E|14|F|G]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|15|8|@]|9|@]|A|$]]]|R|$]]

Because you can't. In C++ language fully-qualified names are only used to refer to existing (i.e. previously declared) entities. They can't be used to introduce new entities.

And you are in fact "reopening" the namespace to declare new entities. If the class <code>Class</code> is later defined as a member of different namespace - it is a completely different class that has nothing to do with the one you declared here.

Once you get to the point of defining the pre-declared class, you don't need to "reopen" the namespace again. You can define it in the global namespace (or any namespace enclosing your <code>Namespace</code>) as

<pre><code>class Namespace::Class {
 /* whatever */
};
</code></pre>

Since you are referring to an entity that has already been declared in namespace <code>Namespace</code>, you can use qualified name <code>Namespace::Class</code>.

blocks|key|3282115|text|你得到了正确的答案，让我试着重新措辞：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3282116|class+Namespace::Class;|offset|length|style|CODE|3282117|3282118|为什么我必须这样做？|blockquote|3282119|3282120|3282121|你必须这样做，因为术语Namespace::Class告诉编译器：|3282122|3282123|+...OK，编译器。找到名为Namespace的名称空间，并在其中引用名为Class的类。|3282124|3282125|3282126|但是编译器不知道您在说什么，因为它不知道任何名为Namespace的名称空间。即使有一个名为Namespace的名称空间，如下所示：|3282127|namespace+Namespace
{
};

class+Namespace::Class;|code-block|syntax|javascript|3282128|它仍然不能工作，因为您不能在名称空间内从该名称空间外部声明一个类。您必须在名称空间中。|3282129|因此，您实际上可以在名称空间中向前声明一个类。只需执行以下操作：|3282130|namespace+Namespace
{
++++class+Class;
};|3282131|entityMap^0|0|0|N|0|0|0|0|0|B|G|0|0|0|0|0|O|9|1A|9|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|19|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1A|8|@$D|1B|E|1C|F|G]]|9|@]|A|$]]|$1|H|3|-4|5|6|7|1D|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|1E|8|@]|9|@]|A|$]]|$1|L|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|1G|8|@]|9|@]|A|$]]|$1|N|3|O|5|6|7|1H|8|@$D|1I|E|1J|F|G]]|9|@]|A|$]]|$1|P|3|-4|5|6|7|1K|8|@]|9|@]|A|$]]|$1|Q|3|R|5|K|7|1L|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1M|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1O|8|@$D|1P|E|1Q|F|G]|$D|1R|E|1S|F|G]]|9|@]|A|$]]|$1|W|3|X|5|Y|7|1T|8|@]|9|@]|A|$Z|10]]|$1|11|3|12|5|6|7|1U|8|@]|9|@]|A|$]]|$1|13|3|14|5|6|7|1V|8|@]|9|@]|A|$]]|$1|15|3|16|5|Y|7|1W|8|@]|9|@]|A|$Z|10]]|$1|17|3|-4|5|6|7|1X|8|@]|9|@]|A|$]]]|18|$]]

You're getting correct answers, let me just try re-wording:

<code>class Namespace::Class;</code>

<blockquote>
 Why do I have to do this?
</blockquote>

You have to do this because the term <code>Namespace::Class</code> is telling the compiler:

<blockquote>
 ...OK, compiler. Go find the
 namespace named Namespace, and within
 that refer to the class named Class.
</blockquote>

But the compiler doesn't know what you're talking about because it doesn't know any namespace named <code>Namespace</code>. Even if there were a namespace named <code>Namespace</code>, as in:

<pre><code>namespace Namespace
{
};

class Namespace::Class;
</code></pre>

it still wouldn't work, because you can't declare a class within a namespace from outside that namespace. You have to be in the namespace.

So, you can in fact forward declare a class within a namespace. Just do this:

<pre><code>namespace Namespace
{
 class Class;
};
</code></pre>

blocks|key|318171|text|我想这也是出于同样的原因，你不能像这样一次性声明嵌套的命名空间：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|318172|namespace+Company::Communications::Sockets+{
}|code-block|syntax|javascript|318173|你必须这样做：|318174|namespace+Company+{
++namespace+Communications+{
++++namespace+Sockets+{
++++}
++}
}|318175|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

I suppose it's for the same reason you cannot declare nested namespaces in one go like this:

<pre><code>namespace Company::Communications::Sockets {
}
</code></pre>

and you have to do this:

<pre><code>namespace Company {
 namespace Communications {
 namespace Sockets {
 }
 }
}
</code></pre>

blocks|key|3282167|text|关于不允许这样做的理由，有很多很好的答案。我只想提供令人厌烦的标准条款，这是明确禁止的。这也适用于C%2B%2B17+(n4659)。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3282168|有问题的段落是[class.name]/2|offset|length|3282169|3282170|只包含类键标识符的声明；可以是当前作用域中的名称的重新声明，也可以是标识符作为类名的向前声明。它将类名引入当前作用域。|blockquote|3282171|3282172|3282173|上面定义了什么构成了转发声明(或类的redclaration+)。本质上，它必须是class+identifier;、struct+identifier;或union+identifier;之一，其中identifer是[lex.name]中的常见词法定义|style|CODE|3282174|数字标识符:标识符-非数字标识符标识符-非数字标识符数字标识符-非数字:非数字通用字符名称非数字:a+b+c+d+e+f+g+h+i+j+k+l+m+n+o+p+q+r+s+t+u+v+w+x+y+z+A+B+C+D+E+F+G+H+i+J+K+L+M+N+O+P+Q+R+S+T+U+V+W+X+Y+Z_+|3282175|之一:0+1+2+3+4+5+6+7+8+9之一|3282176|3282177|3282178|它是我们都熟悉的通用方案[a-zA-Z_][a-zA-Z0-9_]*的产物。正如您所看到的，这排除了class+foo::bar;作为有效的转发声明的可能性，因为foo::bar不是标识符。这是一个完全限定的名称，一些不同的东西。|3282179|entityMap|0|LINK|mutability|MUTABLE|url|https://timsong-cpp.github.io/cppwp/n4659/class.name#2|1|https://timsong-cpp.github.io/cppwp/n4659/lex.name^0|0|7|E|0|0|0|0|0|0|15|H|1N|I|26|H|32|A|1|0|0|0|0|0|C|M|1E|F|29|8|0^^$0|@$1|2|3|4|5|6|7|17|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|18|8|@]|9|@$D|19|E|1A|1|1B]]|A|$]]|$1|F|3|-4|5|6|7|1C|8|@]|9|@]|A|$]]|$1|G|3|H|5|I|7|1D|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|1E|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|1F|8|@]|9|@]|A|$]]|$1|L|3|M|5|6|7|1G|8|@$D|1H|E|1I|N|O]|$D|1J|E|1K|N|O]|$D|1L|E|1M|N|O]]|9|@$D|1N|E|1O|1|1P]]|A|$]]|$1|P|3|Q|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|R|3|S|5|I|7|1R|8|@]|9|@]|A|$]]|$1|T|3|-4|5|6|7|1S|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1T|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1U|8|@$D|1V|E|1W|N|O]|$D|1X|E|1Y|N|O]|$D|1Z|E|20|N|O]]|9|@]|A|$]]|$1|X|3|-4|5|6|7|21|8|@]|9|@]|A|$]]]|Y|$Z|$5|10|11|12|A|$13|14]]|15|$5|10|11|12|A|$13|16]]]]

There's a lot of excellent answers about the rationale involved in disallowing it. I just want to provide the boring standardese clause the specifically prohibits it. This holds true for C++17 (n4659).

The paragraph in question is <a href="https://timsong-cpp.github.io/cppwp/n4659/class.name#2" rel="nofollow noreferrer">[class.name]/2</a>:

<blockquote>
 A declaration consisting solely of class-key identifier; is either a
 redeclaration of the name in the current scope or a forward
 declaration of the identifier as a class name. It introduces the class
 name into the current scope.
</blockquote>

The above defines what constitutes a forward declaration (or redclaration of a class). Essentially, it must be one of <code>class identifier;</code>, <code>struct identifier;</code> or <code>union identifier;</code> where identifer is the common lexical definition in <a href="https://timsong-cpp.github.io/cppwp/n4659/lex.name" rel="nofollow noreferrer">[lex.name]</a>:

<blockquote>
<pre><code>identifier:
 identifier-nondigit
 identifier identifier-nondigit
 identifier digit
identifier-nondigit:
 nondigit
 universal-character-name
nondigit: one of
 a b c d e f g h i j k l m
 n o p q r s t u v w x y z
 A B C D E F G H I J K L M
 N O P Q R S T U V W X Y Z _
digit: one of
 0 1 2 3 4 5 6 7 8 9
</code></pre>
</blockquote>

Which is the production of the common scheme <code>[a-zA-Z_][a-zA-Z0-9_]*</code> we are all familiar with. As you can see, this precludes <code>class foo::bar;</code> from being a valid forward declaration, because <code>foo::bar</code> is not an identifier. It's a fully qualified name, something different.

blocks|key|322995|text|不清楚一个正向声明变量的类型到底是什么。转发声明class+Namespace::Class;可能意味着|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|322996|namespace+Namespace+{
++class+Class;
}|code-block|syntax|javascript|322997|或|322998|class+Namespace+{
public:
++class+Class;
};|322999|entityMap^0|O|N|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

It would not be clear what a forward declared variable's type actually is. The forward declaration <code>class Namespace::Class;</code> could mean

<pre><code>namespace Namespace {
 class Class;
}
</code></pre>

or

<pre><code>class Namespace {
public:
 class Class;
};
</code></pre>

<pre><code>class Namespace::Class;
</code></pre>

Why do I have to do this?:

<pre><code>namespace Namespace {
 class Class;
}
</code></pre>

Using VC++ 8.0, the compiler issues:

<blockquote>
 error C2653: 'Namespace' : is not a class or namespace name
</blockquote>

I assume that the problem here is that the compiler cannot tell whether <code>Namespace</code> is a class or a namespace? But why does this matter since it's just a forward declaration?

Is there another way to forward-declare a class defined in some namespace? The syntax above feels like I'm "reopening" the namespace and extending its definition. What if <code>Class</code> were not actually defined in <code>Namespace</code>? Would this result in an error at some point?

Why can't I forward-declare a class in a namespace using double colons?

编译器

class Namespace::Class;为什么我必须这样做？：namespace Namespace {    class Class;}使用VC++ 8.0时，编译器会出现以下问题：错误名称空间：‘C2653’：不是类或命名空间名称我假设这里的问题是编译器无法判断Namespace是一个类还是一个名称空间？但是...

问为什么我不能在命名空间中使用双冒号向前声明一个类？
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问为什么我不能在命名空间中使用双冒号向前声明一个类？
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