问题描述
我有std :: vector< Base * base;
我想做类似的事情：
std :: for_each（bases .begin（），bases.end（），operator delete）;
是否可以不编写适配器？有没有更好的办法？
有现有适配器吗？
谢谢，
Daniel。
-
Daniel Pitts的'科技博客：< http://virtualinfinity.net/wordpress/>
保持对动态对象的简单指针是太麻烦了。
智能指针是最好的方法。 boost :: shared_ptr是合适的：
#include< boost / shared_prt.hpp>
/ * ... * /
typedef boost :: shared_ptr< BaseBasePtr;
std :: vector< BasePtrbases;
这个和其他问题神奇地消失了......
Ali
出了什么问题：
for（std :: vector< Base *> :: iterator i = bases.begin（）; i！=
bases.end（）; ++ i）
delete * i;
它只有两条线;而且很明显发生了什么。我不认为
你能打败它。如果你经常这样做，你可以写一个实用函数，如：
模板< class Tvoid delete_all（T& cont）{
typedef typename T :: iterator iter_t;
for（iter_t i（cont.begin（））; i！= cont.end（）; ++ i）
删除* i;
cont.clear（）;
}
示例：
struct A {...};
void f（）{
vector< A * x;
list< A * y;
set< A * z;
x.push_back（new A）;
x.push_back（新A）;
y.push_back（新A）;
y.push_back（新A）;
z.insert（新A）;
z.insert（新A）;
delete_all（x）;
delete_all（y）;
delete_all（z）;
}
Jason
看起来不是这样......你最好写自己的my_delete：
模板< class T>
struct my_delete：std :: unary_function< T，void>
{
void operator（）（T * t）
{
删除t;
}
};
这一个：
std :: for_each（bases.begin（），bases.end（），std :: ptr_fun（operator delete））;
编译，但不起作用（dtor未被调用）...
-
William
Underdogging：
在给定的情况下，几乎总是与弱者站在一起的倾向。这种特性的消费者表达是购买不太成功的悲伤，悲伤，或者是失败的产品：我知道这些维也纳弗兰克人棒棒糖是心脏衰竭，但他们太伤心了
查看所有其他雅皮士食品我只需要
就可以买到它们。
- Douglas Coupland，X世代：加速的故事
文化< br />
I have std::vector<Base *bases;
I''d like to do something like:
std::for_each(bases.begin(), bases.end(), operator delete);
Is it possible without writing an adapter? Is there a better way? Is
there an existing adapter?
Thanks,
Daniel.
--
Daniel Pitts'' Tech Blog: <http://virtualinfinity.net/wordpress/>
Keeping plain pointers to dynamic objects is way too much trouble.
Smart pointers is the best way. boost::shared_ptr is suitable:
#include <boost/shared_prt.hpp>
/* ... */
typedef boost::shared_ptr<BaseBasePtr;
std::vector<BasePtrbases;
This and other problems magically vanish...
Ali
What''s wrong with:
for (std::vector<Base *>::iterator i = bases.begin(); i !=
bases.end(); ++ i)
delete *i;
It''s only two lines; and it''s clear what is happening. I don''t think
you''ll be able to beat that. If it''s something you do frequently you
could write a utility function like:
template <class Tvoid delete_all (T &cont) {
typedef typename T::iterator iter_t;
for (iter_t i(cont.begin()); i != cont.end(); ++ i)
delete *i;
cont.clear();
}
Example:
struct A { ... };
void f () {
vector<A *x;
list<A *y;
set<A *z;
x.push_back(new A);
x.push_back(new A);
y.push_back(new A);
y.push_back(new A);
z.insert(new A);
z.insert(new A);
delete_all(x);
delete_all(y);
delete_all(z);
}
Jason
Doesn''t seem so... You''d better write your own my_delete:
template <class T>
struct my_delete : std::unary_function<T, void>
{
void operator()(T* t)
{
delete t;
}
};
This one:
std::for_each(bases.begin(), bases.end(), std::ptr_fun(operator delete));
compiles, but doesn''t work(dtor is not called)...
--
William
http://williamxu.net9.org
Underdogging:
The tendency to almost invariably side with the underdog in a
given situation. The consumer expression of this trait is the
purchasing of less successful, "sad," or failing products: "I know
these Vienna franks are heart failure on a stick, but they were so sad
looking up against all the other yuppie food items that I just had to
buy them."
-- Douglas Coupland, "Generation X: Tales for an Accelerated
Culture"
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