一个简单的c ++问题...... [英] A simple c++ question...
问题描述
我有几节课...
a类：std :: list< baseclass>
{
int keya;
};
class b：std :: list< a>
{
int keyb;
};
class c：std :: list< b>
{
int keyc;
};
我想在每个级别弄清楚列表的大小......
如果我有：
int main（int argc，char * argv []）
{
cx;
cout<< c类列表的大小是 << x.size（）<< endl;
cout<< b类列表的大小是 << ???? << endl;
cout<< 班级列表的大小是 << ???? <<< endl;
}
------------------------------- ------------------------------------------
想一想......
使用std :: list的继承可能不是最好的想法..
或许：
类a：std :: list< baseclass>
{
int keya;
std :: list< sometype> ; baseclasslist;
};
class b
{
int keyb;
std :: list< a> alist;
};
class c
{
int keyc;
std :: list< b> blist;
};
可能是一个更好的实现，但问题仍然存在。
I have a few classes...
class a : std::list<baseclass>
{
int keya;
};
class b : std::list<a>
{
int keyb;
};
class c : std::list<b>
{
int keyc;
};
I''m trying to figure out at each level the size of the lists...
so if I have:
int main(int argc, char *argv[])
{
c x;
cout << "The size of the c class list is " << x.size() << endl;
cout << "The size of the b class list is " << ???? << endl;
cout << "The size of the a class list is " << ???? <<< endl;
}
-------------------------------------------------------------------------
Thinking about it...
Using inheritance of std::list may not have been the best idea..
perhaps:
class a : std::list<baseclass>
{
int keya;
std::list<sometype> baseclasslist;
};
class b
{
int keyb;
std::list<a> alist;
};
class c
{
int keyc;
std::list<b> blist;
};
May have been a better implementation, but the question remains.
推荐答案
" JustSomeGuy" <无** @ nottelling.com>写了...
"JustSomeGuy" <no**@nottelling.com> wrote...
我有几节课......
[...]
I have a few classes...
[..]
耐心是一种美德/>
Patience is a virtue
你的问题是你私下里，所以你不能
访问a，b，c的超类中的size（）函数等等。这是'b $ ba工作的一段代码''splain it。
davet。
#include< iostream>
#include< list>
using namespace std;
class baseclass
{
public：
};
class A：public std :: list< int>
{
int keya;
};
B类：public std :: list< A> ;
{
int keyb;
};
class C： public std :: list< B>
{
int keyc;
};
int main（）
{
A a;
B b;
C c;
for（int k = 0; K< 10; k ++）
{
for（int j = 0; j< 10; j ++）
{
for （int i = 0; i< 10; i ++）
{
a.push_back（i）;
}
b.push_back（a）;
}
c.push_back（b）;
}
cout<< 班级列表的大小是 << a.size（）<< endl;
cout<< b类列表的大小是 << b.size（）<< endl;
cout<< c类列表的大小是 << c.size（）<<结束;
返回0;
}
" JustSomeGuy" <无** @ nottelling.com>在消息中写道
news：pxFKc.54262
Your problem was you were inherting privately, so you couldn''t
access the size() function in the superclass of a, b,c etc. Here''s
a working piece of code to ''splain it.
davet.
#include <iostream>
#include <list>
using namespace std;
class baseclass
{
public:
};
class A : public std::list<int>
{
int keya;
};
class B : public std::list<A>
{
int keyb;
};
class C : public std::list<B>
{
int keyc;
};
int main()
{
A a;
B b;
C c;
for (int k=0; k<10; k++)
{
for (int j=0; j<10 ; j++)
{
for (int i=0; i< 10; i++)
{
a.push_back(i);
}
b.push_back(a);
}
c.push_back(b);
}
cout << "The size of the a class list is " << a.size() << endl;
cout << "The size of the b class list is " << b.size() << endl;
cout << "The size of the c class list is " << c.size() << endl;
return 0;
}
"JustSomeGuy" <no**@nottelling.com> wrote in message
news:pxFKc.54262
Mr4.2978@pd7tw1no ...
Mr4.2978@pd7tw1no...
我有几节课.. 。
类a：std :: list< baseclass>
{keya;
};
b类：std： ：list< a>
{key int;
};
类c：std :: list< b>
{
int keyc;
};
我想在每个级别弄清楚列表的大小......
所以如果我有：
int main（int argc，char * argv []）
{
cx;
cout<< c类列表的大小是 << x.size（）<< endl;
cout<< b类列表的大小是 << ???? << endl;
cout<< 班级列表的大小是 << ???? <<< endl;
}
--------------------------------------- ----------------------------------
想一想......也许：
类a：std :: list< baseclass>
{
int keya;
std :: list< sometype> baseclasslist;
};
class b
{
int keyb;
std :: list< a> alist;
};
class c
{
int keyc;
std :: list< b> blist;
};
可能是一个更好的实现，但问题仍然存在。
I have a few classes...
class a : std::list<baseclass>
{
int keya;
};
class b : std::list<a>
{
int keyb;
};
class c : std::list<b>
{
int keyc;
};
I''m trying to figure out at each level the size of the lists...
so if I have:
int main(int argc, char *argv[])
{
c x;
cout << "The size of the c class list is " << x.size() << endl;
cout << "The size of the b class list is " << ???? << endl;
cout << "The size of the a class list is " << ???? <<< endl;
}
-------------------------------------------------------------------------
Thinking about it...
Using inheritance of std::list may not have been the best idea..
perhaps:
class a : std::list<baseclass>
{
int keya;
std::list<sometype> baseclasslist;
};
class b
{
int keyb;
std::list<a> alist;
};
class c
{
int keyc;
std::list<b> blist;
};
May have been a better implementation, but the question remains.
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