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
耐心是一种美德/>
Patience is a virtue
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