使用声明问题 [英] using declaration question
问题描述
为什么我不能在使用声明中使用类析构函数：
使用MyClass :: ~MyClass;
？
-
Christopher Benson-Manica |我*应该*知道我在说什么 - 如果我
ataru（at）cyberspace.org |不，我需要知道。火焰欢迎。
" Christopher Benson-Manica" <在*** @ nospam.cyberspace.org>写在
消息新闻：c1 ********** @ chessie.cirr.com ...为什么我不能上课使用声明中的析构函数：
使用MyClass ::〜MyClass;
有趣的问题！它似乎被语法所允许;然而，由于某些其他原因，它可能是非法的。
VC7.1是我尝试过的唯一接受它的编译器。
Jonathan
Christopher Benson-Manica写道：
为什么我不能用使用声明中的类析构函数：
使用MyClass ::〜MyClass;
因为析构函数（如构造函数）没有名称，并使用
声明需要一个名字。
-
Pete Becker
Dinkumware，Ltd。（ http://www.dinkumware.com ）
" Christopher Benson-Manica" <在*** @ nospam.cyberspace.org>在留言中写道
新闻：c1 ********** @ chessie.cirr.com ...为什么我不能上课使用声明中的析构函数：
使用MyClass ::〜MyClass;
============= ===================================== ============= =====
ISO / IEC 14882：1998（E）
7.3.3使用声明
4 A使用* -declaration作为成员* -declaration将
引用给所定义类的基类成员，
指的是匿名联盟的成员，该联盟是所定义类的
基类的成员，或者应引用
枚举器作为枚举类型的成员一个基础
类定义的类。 [例如：
class C {
int g（）;
};
类D2：公共B {
使用B :: f; // OK：B是使用B :: e的D2
的基础; //好的：e是使用B :: x的基数B的枚举器
; // OK：x是使用C :: g的基数B
的联合成员; //错误：C不是D2的基础
};
- 末端示例] [注意：因为构造函数和析构函数不是
有名称，使用* -declaration不能引用构造函数<< =====
或基类的析构函数。由于名称查找找不到转换函数的成员
模板的特化，
当使用* -declaration指定
$ b $时不考虑它们b转换函数（14.5.2）。 ]如果赋值运算符从基类带入了派生类范围，则派生类的签名为
a copy * -assignment运算符（12.8），使用* -
声明本身不会抑制派生类副本* -assignment运算符的隐式声明
;来自基类的副本* -assignment
运算符被派生类的
隐式* -declared copy * -assignment运算符隐藏或覆盖，
如下所述。
================================== ================ ==================
-Mike
Why can''t I use a class destructor in a using declaration:
using MyClass::~MyClass;
?
--
Christopher Benson-Manica | I *should* know what I''m talking about - if I
ataru(at)cyberspace.org | don''t, I need to know. Flames welcome.
"Christopher Benson-Manica" <at***@nospam.cyberspace.org> wrote in
message news:c1**********@chessie.cirr.com...Why can''t I use a class destructor in a using declaration:
using MyClass::~MyClass;
Interesting question! It seems to be allowed by the grammar; it may be
illegal for some other reason, however.
VC7.1 is the only compiler I''ve tried which accepts it.
Jonathan
Christopher Benson-Manica wrote:
Why can''t I use a class destructor in a using declaration:
using MyClass::~MyClass;
Because destructors, like constructors, do not have names, and a using
declaration requires a name.
--
Pete Becker
Dinkumware, Ltd. (http://www.dinkumware.com)
"Christopher Benson-Manica" <at***@nospam.cyberspace.org> wrote in message
news:c1**********@chessie.cirr.com...Why can''t I use a class destructor in a using declaration:
using MyClass::~MyClass;
================================================== ==================
ISO/IEC 14882:1998(E)
7.3.3 The using declaration
4 A using*-declaration used as a member*-declaration shall refer
to a member of a base class of the class being defined, shall
refer to a member of an anonymous union that is a member of a
base class of the class being defined, or shall refer to an
enumerator for an enumeration type that is a member of a base
class of the class being defined. [Example:
class C {
int g();
};
class D2 : public B {
using B::f; // OK: B is a base of D2
using B::e; // OK: e is an enumerator of base B
using B::x; // OK: x is a union member of base B
using C::g; // error: C isn?t a base of D2
};
--end example] [Note: since constructors and destructors do not
have names, a using*-declaration cannot refer to a constructor <<=====
or a destructor for a base class. Since specializations of member
templates for conversion functions are not found by name lookup,
they are not considered when a using*-declaration specifies a
conversion function (14.5.2). ] If an assignment operator brought
from a base class into a derived class scope has the signature of
a copy*-assignment operator for the derived class (12.8), the using*-
declaration does not by itself suppress the implicit declaration
of the derived class copy*-assignment operator; the copy*-assignment
operator from the base class is hidden or overridden by the
implicitly*-declared copy*-assignment operator of the derived class,
as described below.
================================================== ==================
-Mike
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