从STL容器派生 [英] Deriving from STL containers
问题描述
HI
如果我定义类DoubleMap，那么
struct DoubleMap：public std :: map< std :: string，double> {} ;
调用std :: map成员函数是否有任何开销？
此外是STL容器析构函数虚拟>
N
BTW ..我知道我可以做一个typedef
所以让我重新定义DoubleMap以澄清我的问题。
模板< typename键>
struct DoubleMap：public std :: map< Key，双> {};
Nind ... @ yahoo.co.uk写道：
HI
如果我定义DoubleMap类，那么
struct DoubleMap：public std :: map< std :: string，double> {};
调用std :: map成员函数是否有任何开销？
此外是STL容器析构函数虚拟>
N
再次抱歉.....假设我已经投入必要的
构造函数。
是
模板< typename键>
struct DoubleMap：public std :: map< Key，double> {
DoubleMap（）：std :: map< Ky，double>（）{}
};
Ni ***** @ yahoo.co.uk 写道：
BTW ..我知道我可以做一个typedef
所以让我重新定义DoubleMap来澄清我的问题。
模板< typename键>
struct DoubleMap：public std： ：map< Key，double> {};
Nind ... @ yahoo.co.uk写道：
HI
如果我定义DoubleMap类，那么
struct DoubleMap：public std :: map< std :: string，double> {};
调用std :: map成员函数是否有任何开销？
Moreo ver是STL容器析构函数虚拟>
N
Ni ***** @ yahoo.co.uk 写道：
BTW ..我知道我可以做一个typedef
所以让我重新定义DoubleMap以澄清我的问题。
template< typename Key>
struct DoubleMap：public std :: map< Key，double> {};
Nind ... @ yahoo.co.uk写道：
哎呀：（a）请注意，在这个群组中，顶级帖子不赞成所有常客。
$ br />
> HI
如果我定义类DoubleMap，那么
struct DoubleMap：public std :: map< std :: string，double> {};
调用std :: map成员函数有什么开销吗？
No.
>此外是STL容器析构函数虚拟>
No.这意味着你不应该使用指向STL容器的指针
多态。
另请注意，当您拥有带签名的
函数时，公共继承可以产生有趣的结果
STL_container< Tdo_something（STL_container< Tconst&）;
在这种情况下，你的派生类会匹配，但返回类型将是
可能不是你所期望的。请注意，私有继承不会涉及任何这种棘手的问题。
> >
N
哎呀：另外：请不要引用签名。
Best
Kai-Uwe Bux
HI
If I define the class DoubleMap such that
struct DoubleMap : public std::map<std::string, double>{};
Is there any overhead in calling std::map member functions ?
Moreover are STL containers destructors virtual >
N
解决方案BTW .. I know I can do a typedef
so let me re-define DoubleMap to clarify my question .
template<typename Key>
struct DoubleMap : public std::map<Key, double>{};
Nind...@yahoo.co.uk wrote:HI
If I define the class DoubleMap such that
struct DoubleMap : public std::map<std::string, double>{};
Is there any overhead in calling std::map member functions ?
Moreover are STL containers destructors virtual >
N
again sorry ..... lets suppose I have put in the necessary
constructors.
the is
template<typename Key>
struct DoubleMap : public std::map<Key, double>{
DoubleMap():std::map<Ky,double>(){}
};
Ni*****@yahoo.co.uk wrote:BTW .. I know I can do a typedef
so let me re-define DoubleMap to clarify my question .
template<typename Key>
struct DoubleMap : public std::map<Key, double>{};
Nind...@yahoo.co.uk wrote:HI
If I define the class DoubleMap such that
struct DoubleMap : public std::map<std::string, double>{};
Is there any overhead in calling std::map member functions ?
Moreover are STL containers destructors virtual >
N
Ni*****@yahoo.co.uk wrote:
BTW .. I know I can do a typedef
so let me re-define DoubleMap to clarify my question .
template<typename Key>
struct DoubleMap : public std::map<Key, double>{};
Nind...@yahoo.co.uk wrote:Oops: (a) please note that top-posting is frowned upon in this group by
virtually all regulars.
>HI
If I define the class DoubleMap such that
struct DoubleMap : public std::map<std::string, double>{};
Is there any overhead in calling std::map member functions ?
No.
>Moreover are STL containers destructors virtual >
No. That means you should not use pointers to STL containers
polymorphically.
Also note that public inheritance can yield funny results when you have
functions with signature
STL_container<Tdo_something ( STL_container<Tconst & );
In this case, your derived classes will match, but the return type will
probably not be what you expect. Note that private inheritance does not
involve any of this trickyness.
>>
N
Oops: Also: please do not quote signatures.
Best
Kai-Uwe Bux
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