关于成员变量初始化和默认构造函数问题 [英] About member variable initilization and default constructor issues
问题描述
在Inside C ++对象模型中，Lippman说有四种情况，如果构造函数不存在，
compile将合成一个默认构造函数来初始化成员
变量：
1.有虚拟功能;
2.虚拟继承;
3带有显式默认构造函数的.base类;
4.使用显式默认构造函数的成员对象。
例如
A级{
公开：
int i;
A * p;
}
int main（）
{
A a;
if（ai == 0 || ap == 0）//做点什么
}
对于上述情况，行为未定义。如果我们将A类修改为
有四个characeristics中的任何一个，那么成员变量
将在默认构造函数中初始化，由
$ b表示。 $ b编译器，好像
A（）：i（0），p（0）{}在类中定义。
但我试试这在不同的编译器中，产生不同的结果。我的
问题是：Lippman教授ISO标准，还是编译器 -
依赖？谢谢
In Inside C++ object Model, Lippman said there are four cases in which
compile will sythesize a default constructor to initialize the member
variables if the constructor is absent:
1. there is a virtual function;
2. virtual inheritance;
3.base class with explicit default constructor;
4.member object with explicit default constructor.
e.g.
class A {
public:
int i;
A* p;
}
int main()
{
A a;
if(a.i ==0 || a.p == 0) //do something
}
The behavior is undefined for the above case. If we modify class A to
have any one of the four characeristics, then the member variables
will be initialized in the default constructor sythesysized by the
compiler, as though
A():i(0),p(0){} is defined in the class.
But I try this in different compilers, and yield different results. My
question is : Is Lippman teaching ISO standard, or compiler-
dependent? Thanks
推荐答案
JosephLee写道：
JosephLee wrote:
In C ++对象模型，Lippman说有四种情况，如果构造函数不存在，
compile会合成一个默认构造函数来初始化成员
变量：
1.有一个虚函数;
2.虚拟继承;
3.base类，带显式默认构造函数;
4.使用显式默认构造函数构成对象。
例如
class A {
public：
int i;
A * p;
}
int main（）
{
A a;
if（ai == 0 | | ap == 0）//做点什么
}
对于上述情况，行为未定义。如果我们将A类修改为
有四个characeristics中的任何一个，那么成员变量
将在默认构造函数中初始化，由
$ b表示。 $ b编译器，好像
A（）：i（0），p（0）{}在类中定义。
但我试试这在不同的编译器中，产生不同的结果。我的
问题是：Lippman教授ISO标准，还是编译器 -
依赖？谢谢
In Inside C++ object Model, Lippman said there are four cases in which
compile will sythesize a default constructor to initialize the member
variables if the constructor is absent:
1. there is a virtual function;
2. virtual inheritance;
3.base class with explicit default constructor;
4.member object with explicit default constructor.
e.g.
class A {
public:
int i;
A* p;
}
int main()
{
A a;
if(a.i ==0 || a.p == 0) //do something
}
The behavior is undefined for the above case. If we modify class A to
have any one of the four characeristics, then the member variables
will be initialized in the default constructor sythesysized by the
compiler, as though
A():i(0),p(0){} is defined in the class.
But I try this in different compilers, and yield different results. My
question is : Is Lippman teaching ISO standard, or compiler-
dependent? Thanks
标准说如果构造函数隐含地定义了
并且该类没有虚函数或虚基，那么构造函数是微不足道的，所有直接
基类都有琐碎的c-tors，所有非静态数据成员也都有b
琐碎的c-tors。所以，转过来，你会得到Lippman所说的。
V
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The Standard says that the constructor is trivial if it''s implicitly
defined and the class no virtual functions or virtual bases, all direct
base classes have trivial c-tors, all non-static data members also have
trivial c-tors. So, turn that around and you get what Lippman says.
V
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嗨
Victor Bazarov写道：
Hi
Victor Bazarov wrote:
JosephLee写道：
JosephLee wrote:
>在Inside C ++对象Model中，Lippman说有四种情况，如果构造函数不存在，
compile会合成一个默认构造函数来初始化成员
变量：
1。有一个虚拟功能;
2。虚继承;
3.base类具有显式默认构造函数;
4.member具有显式默认构造函数的对象。
>In Inside C++ object Model, Lippman said there are four cases in which
compile will sythesize a default constructor to initialize the member
variables if the constructor is absent:
1. there is a virtual function;
2. virtual inheritance;
3.base class with explicit default constructor;
4.member object with explicit default constructor.
[...]
[...]
> ;如果我们将A类修改为具有四个characeristics中的任何一个，那么成员变量
将在由
编译器sythesys化的默认构造函数中初始化，就好像
A （）：i（0），p（0）{}在类中定义。
>If we modify class A to
have any one of the four characeristics, then the member variables
will be initialized in the default constructor sythesysized by the
compiler, as though
A():i(0),p(0){} is defined in the class.
标准说如果构造函数隐含地定义为
并且类没有虚函数或虚拟基，那么构造函数是微不足道的直接
基类有琐碎的c-tors，所有非静态数据成员也都有b
琐碎的c-tors。所以，转过来，你会得到Lippman说的话。
The Standard says that the constructor is trivial if it''s implicitly
defined and the class no virtual functions or virtual bases, all direct
base classes have trivial c-tors, all non-static data members also have
trivial c-tors. So, turn that around and you get what Lippman says.
我不同意。首先，如果1-4不满足则不正确那么
将没有合成的默认构造函数：
如果有的话没有用户声明的类X的构造函数，默认构造函数
是隐式声明的。
"类的隐式声明的默认构造函数在用于创建其类类型的对象时，隐式地定义了
。
隐式定义的默认构造函数执行初始化的集合
将由用户编写的默认构造函数执行
for那个带有空的mem-initializer-list和一个空函数的类
body"
其次，后一个短语解释了在上述两个案例中，
程序的行为就好像有一个用户定义的构造函数
A（）{}
这意味着我和p都不会被初始化。
Markus
I disagree. Firstly, it''s not correct that if 1-4 is not satisfied then
there will be no default-constructor synthesized:
"If there is no user-declared constructor for class X, a default constructor
is implicitly declared."
"An implicitly-declared default constructor for a class is implicitly
defined when it is used to create an object of its class type. The
implicitly-defined default constructor performs the set of initializations
of the class that would be performed by a user-written default constructor
for that class with an empty mem-initializer-list and an empty function
body"
Secondly, the latter phrase explains that in both of the above cases, the
program will behave as if there was a user-defined constructor
A() {}
This means that neither i nor p will be initialized.
Markus
Markus Moll写道：
Markus Moll wrote:
嗨
Victor Bazarov写道：
Hi
Victor Bazarov wrote:
> JosephLee写道：
>JosephLee wrote:
>>在Inside C ++对象模型中，Lippman说
中有四种情况，编译将sythesize默认构造函数初始化
会员va如果构造函数不存在则会出现问题：
1。有一个虚拟功能;
2。虚继承;
3.base类具有显式默认构造函数;
4.member具有显式默认构造函数的对象。
>>In Inside C++ object Model, Lippman said there are four cases in
which compile will sythesize a default constructor to initialize
the member variables if the constructor is absent:
1. there is a virtual function;
2. virtual inheritance;
3.base class with explicit default constructor;
4.member object with explicit default constructor.
[...]
[...]
> ;>如果我们修改A类到
有四个characeristics中的任何一个，那么成员变量
将在
编译器sythesysized的默认构造函数中初始化，就好像 A（）：i（0），p（0）{}在类中定义。
>>If we modify class A to
have any one of the four characeristics, then the member variables
will be initialized in the default constructor sythesysized by the
compiler, as though
A():i(0),p(0){} is defined in the class.
标准说如果构造函数被隐式定义并且类没有虚函数或虚函数，所有
直接基类都有，那么构造函数是微不足道的琐碎的c-tors，所有非静态数据成员也有微不足道的c-tors。所以，转过来，你会得到什么
Lippman说。
The Standard says that the constructor is trivial if it''s implicitly
defined and the class no virtual functions or virtual bases, all
direct base classes have trivial c-tors, all non-static data members
also have trivial c-tors. So, turn that around and you get what
Lippman says.
我不同意。首先，如果1-4不满足则不正确
则不会合成默认构造函数：
如果有的话没有用户声明的类X的构造函数，默认声明了一个默认的
构造函数。
一个隐式声明的类的默认构造函数在用于创建其类类型的对象时，隐式地定义了
。
隐式定义的默认构造函数执行该类的初始化，该初始化将由
用户编写的默认构造函数执行。那个带有空的
mem-initializer-list和一个空函数体的类'
其次，后一个短语解释了在上述两种情况中，
程序的行为就好像有一个用户定义的构造函数
A（）{}
这意味着i和p都不会被初始化。
I disagree. Firstly, it''s not correct that if 1-4 is not satisfied
then there will be no default-constructor synthesized:
"If there is no user-declared constructor for class X, a default
constructor is implicitly declared."
"An implicitly-declared default constructor for a class is implicitly
defined when it is used to create an object of its class type. The
implicitly-defined default constructor performs the set of
initializations of the class that would be performed by a
user-written default constructor for that class with an empty
mem-initializer-list and an empty function body"
Secondly, the latter phrase explains that in both of the above cases,
the program will behave as if there was a user-defined constructor
A() {}
This means that neither i nor p will be initialized.
对。我不确定你在和谁争论。在
类中导致隐式定义的非平凡c-tor的四个特定的
特征与标准中的相同。 Lippman说，如果存在四个中的任何一个，那么将产生c-tor
。 Stadard说，如果这四个人中没有一个存在，那么c-tor是微不足道的。从那个意义上来说，李普曼教授标准。不是吗？
V
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Right. I am not sure with whom you''re arguing. The four specific
traits that cause an implicitly defined non-trivial c-tor in the
class are the same as in the Standard. Lippman says that the c-tor
is generated if any of the four is present. The Stadard says that
the c-tor is trivial if none of the four is present. In that sense
Lippman teaches the Standard. No?
V
--
Please remove capital ''A''s when replying by e-mail
I do not respond to top-posted replies, please don''t ask
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