访问和名称查找，这个代码感到惊讶 [英] access and name lookup, surprised by this code
问题描述
gcc 3.3.1和VC ++ 7.1都编译以下代码。
struct Outer
{
struct Inner
{
int f（）{return c; }
};
私人：
static const int c;
};
const int Outer :: c = 123;
int main（）
{
外部： ：内部o;
std :: cout<< o.f（）<< ''\\ n';;
}
显然我错了，但我认为以上打破了两个不同的规则
C ++。首先不是私人，其次如何内部访问
Outer的成员而没有限定名称（即返回Outer :: c）？
解释我将不胜感激。
john
Both gcc 3.3.1 and VC++ 7.1 compile the following code.
struct Outer
{
struct Inner
{
int f() { return c; }
};
private:
static const int c;
};
const int Outer::c = 123;
int main()
{
Outer::Inner o;
std::cout << o.f() << ''\n'';
}
Obviously I''m wrong but I thought the above breaks two different rules of
C++. Firstly isn''t c private, secondly how can Inner access a member of
Outer without qualifying the name (i.e. return Outer::c)?
Explanations and references to the standard would be appreciated.
john
推荐答案
John Harrison <乔************* @ hotmail.com>在消息中写道
news：c1 ************* @ ID-196037.news.uni-berlin.de ...
"John Harrison" <jo*************@hotmail.com> wrote in message
news:c1*************@ID-196037.news.uni-berlin.de...
gcc 3.3.1和VC ++ 7.1都编译了以下代码。
struct Outer
{
struct Inner
{
int f（）{return c ;私有：
static const int c;
};
const int Outer :: c = 123;
int main（）
{Outer :: Inner o;
std :: cout<< o.f（）<< ''\ n'';
}
显然我错了，但我认为上面打破了两个不同的C ++规则。首先不是私人的，其次如何内部访问外部的成员而不限定名称（即返回外部:: c）？
对标准的解释和参考将不胜感激。
Both gcc 3.3.1 and VC++ 7.1 compile the following code.
struct Outer
{
struct Inner
{
int f() { return c; }
};
private:
static const int c;
};
const int Outer::c = 123;
int main()
{
Outer::Inner o;
std::cout << o.f() << ''\n'';
}
Obviously I''m wrong but I thought the above breaks two different rules of
C++. Firstly isn''t c private, secondly how can Inner access a member of
Outer without qualifying the name (i.e. return Outer::c)?
Explanations and references to the standard would be appreciated.
我对你的第一个问题感到困惑。
但至于你的第二个问题 -
第9.7节/ 1
"除了使用显式指针，引用和对象名称外，嵌套类中的声明
只能使用类型名称，静态成员和调查员来自
封闭类。 [例如：
int x;
int y;
类附上{
public：
int x;
static int s;
class inner {
void f（int i）
{
int a = sizeof（x）; //错误：指的是封闭:: x
x = i; //错误：指定封闭:: x
s = i; //好的：指定封闭:: s
:: x = i; // OK：分配给全局x
y = i; // OK：分配给全局y
}
void g（附上* p，int i）
{
p-> x = i; // OK：指定封闭:: x
}
};
};
Am perplexed about your 1st question.
But as for your second question -
Section 9.7/1
"Except by using explicit pointers, references, and object names, declarations
in a nested class can use only type names, static members, and enumerators from
the enclosing class". [Example:
int x;
int y;
class enclose {
public:
int x;
static int s;
class inner {
void f(int i)
{
int a = sizeof(x); // error: refers to enclose::x
x = i; // error: assign to enclose::x
s = i; // OK: assign to enclose::s
::x = i; // OK: assign to global x
y = i; // OK: assign to global y
}
void g(enclose* p, int i)
{
p->x = i; // OK: assign to enclose::x
}
};
};
" Sharad Kala" <无***************** @ yahoo.com>在消息中写道
news：c1 ************* @ ID-221354.news.uni-berlin.de ...
"Sharad Kala" <no*****************@yahoo.com> wrote in message
news:c1*************@ID-221354.news.uni-berlin.de...
John Harrison <乔************* @ hotmail.com>在消息中写道
新闻：c1 ************* @ ID-196037.news.uni-berlin.de ...
"John Harrison" <jo*************@hotmail.com> wrote in message
news:c1*************@ID-196037.news.uni-berlin.de...
两者都是gcc 3.3。 1和VC ++ 7.1编译以下代码。
struct Outer
结构内部
{
int f（）{return c;私有：
static const int c;
};
const int Outer :: c = 123;
int main（）
{Outer :: Inner o;
std :: cout<< o.f（）<< ''\ n'';
}
显然我错了，但我认为上面打破了两个不同的C ++规则。首先不是私人的，其次如何内部访问外部的成员而不限定名称（即返回外部:: c）？
对标准的解释和参考将不胜感激。
Both gcc 3.3.1 and VC++ 7.1 compile the following code.
struct Outer
{
struct Inner
{
int f() { return c; }
};
private:
static const int c;
};
const int Outer::c = 123;
int main()
{
Outer::Inner o;
std::cout << o.f() << ''\n'';
}
Obviously I''m wrong but I thought the above breaks two different rules of
C++. Firstly isn''t c private, secondly how can Inner access a member of
Outer without qualifying the name (i.e. return Outer::c)?
Explanations and references to the standard would be appreciated.
对你的第一个问题感到困惑。
Am perplexed about your 1st question.
这不仅仅是静态成员的定义： - ）？
Isn''t it just simply a definition for the static member :-) ?
" Sharad Kala" <无***************** @ yahoo.com>在消息中写道
news：c1 ************* @ ID-221354.news.uni-berlin.de ...
"Sharad Kala" <no*****************@yahoo.com> wrote in message
news:c1*************@ID-221354.news.uni-berlin.de...
Sharad Kala <无***************** @ yahoo.com>在消息中写道
新闻：c1 ************* @ ID-221354.news.uni-berlin.de ...
"Sharad Kala" <no*****************@yahoo.com> wrote in message
news:c1*************@ID-221354.news.uni-berlin.de...
约翰哈里森 <乔************* @ hotmail.com>在消息中写道
新闻：c1 ************* @ ID-196037.news.uni-berlin.de ...
"John Harrison" <jo*************@hotmail.com> wrote in message
news:c1*************@ID-196037.news.uni-berlin.de...
两者都是gcc 3.3。 1和VC ++ 7.1编译以下代码。
struct Outer
结构内部
{
int f（）{return c;私有：
static const int c;
};
const int Outer :: c = 123;
int main（）
{Outer :: Inner o;
std :: cout<< o.f（）<< ''\ n'';
}
显然我错了，但我认为以上打破了两个不同规则
的C ++。首先不是私有的，其次如何在没有限定名称的情况下内部访问成员
的外部（即返回Outer :: c）？
对标准的解释和参考将不胜感激。
Both gcc 3.3.1 and VC++ 7.1 compile the following code.
struct Outer
{
struct Inner
{
int f() { return c; }
};
private:
static const int c;
};
const int Outer::c = 123;
int main()
{
Outer::Inner o;
std::cout << o.f() << ''\n'';
}
Obviously I''m wrong but I thought the above breaks two different rules of C++. Firstly isn''t c private, secondly how can Inner access a member of Outer without qualifying the name (i.e. return Outer::c)?
Explanations and references to the standard would be appreciated.
我对你的第一个问题感到困惑。
Am perplexed about your 1st question.
这不仅仅是静态成员的定义:-)？
Isn''t it just simply a definition for the static member :-) ?
我认为有一条规则，内部类没有特殊的
访问它们的封闭类。所以Inner应该被宣布为
Outer的朋友才能访问Outer :: c，至少那是我的想法。
john
I thought there was a rule to the effect that inner classes have no special
access to their enclosing class. So Inner should be declared a friend of
Outer in order to access Outer::c, at least that''s what I thought.
john
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