问题
I have a quick question regarding the use of typedefs for lengthy templates. The crux: I've found myself in something of a pickle—there doesn't seem to be a good place to place typedefs except local to client functions. While there are similar SO questions (see here for example), none seem to address this exactly. Please note that this question doesn't address whether typedefs are desirable in what follows—I've tried to simplify things for expository purposes.
My problem has arisen while working with boost::shared_ptr<T>. Basically, I want to do the following:
#include <boost/shared_ptr.hpp>
typedef boost::shared_ptr<Widget> WidgetPtr;
Placing this typedef in the Widget declaration header seems ugly. There seem to be two considerations here: (i) if Widget itself doesn't make use of shared pointers in its members, we've added an additional include (as we can't forward declare the boost::shared_ptr template class—correct me if I'm wrong?) (ii) if we want to make use of this typedef during the declaration of another class (call that class Foo) we violate best practices by including Widget.h instead of simply forward declaring Widget or including WidgetFwd.h... unless this typedef is duplicated in the latter. Furthermore, it doesn't seem make sense to typedef boost::shared_ptr<Widget> during the declaration of Widget itself—we seem to be mixing Widget's declaration with an anticipation of how clients will make use of the Widget interface.
Okay, so that's bad, but this is worse: if I don't attempt some combination of the above I end up with duplicate typedefs in client code, which yields inconsistency (and hence, likely, error)—the whole point being that given Widget, a WidgetPtr typedef should act as a type in its own right. Example: we don't want Foo to make use of one WidgetPtr, a typedef of boost::shared_ptr, while Bar is using WidgetPtr as a typedef for std::auto_ptr.
Another method (and one of the few that I've seen mentioned in online discussion) would be to make the typedef a public member of Widget and then use Widget::Ptr:
class Widget {
// ...
public:
typedef boost::shared_ptr<Widget> Ptr;
};
Again, I don't like this as (i) it suggests that the pointer type is somehow a member of the class and (ii) it leads to a wonky interface. Worse still: since every class that I write can potentially be pointed to using smart pointers, I end up chasing the imaginary client's tail. Ugly, ugly, ugly.
As it stands, I've removed the typedefs from this codebase (as they led to serious confusion, duplication) and re-introduced them locally in selected functions. Here again there's a problem with inconsistent usage but it's not quite as severe.
The only other solution I can think of—and again I'm not sure whether this is considered good practice—would be to have a utilities header in which the typedefs are placed, potentially within their own namespace. In this header we'd include and be done with it.
Am I missing something obvious or is this just plain tricky?
PS—Apologies for the length of the above; I couldn't find a simpler way of fully expressing the problem.
回答1:
I don't like a library dictating the use of a particular smart pointer, but I tolerate it if it is necessary.
If you wish to force users to always use a shared_ptr to manipulate a widget, it's impossible, so don't even bother trying.
On the other hand, if you have a method from Widget which returns a boost::shared_ptr<Widget>, then providing a (sane) typedef might simplify the client code.
I would therefore promote the use of an inner typedef:
class Widget
{
public:
typedef boost::shared_ptr<Widget> Ptr;
Ptr AccessFirstChild();
}; // class Widget
in which case it's perfectly okay to #include the necessary headers.
回答2:
Furthermore, it doesn't seem make sense to typedef boost::shared_ptr during the declaration of Widget itself—we seem to be mixing Widget's declaration with an anticipation of how clients will make use of the Widget interface.
First of all, this is not at all wrong - after all, the means of how the clients will (and can) use the interface is part of the interface itself; and for C++, not being garbage-collected, memory management of objects is a rather crucial part of their interface.
So there are two cases. In one case, the Widget would anticipate it would be used through a shared pointer. This would mean that eg. child widgets obtained from a widget are returned as shared_ptrs, everywidget created has it shared_ptr and so on. It would be totally legitimate to typedef WidgetPtr in the same header as Widget.
In the second case, Widgets would expect to be managed eg. by ordinary new and delete. The clients can use shared_ptrs in special cases, but nothing says eg. a printer dialogue routine can't use auto_ptr instead. The clients have to be prepared that if wptr is a shared_ptr, the line
shared_ptr<Widget> w2(wptr->firstChild()->parent());
leads to a disaster.
Your question seems to indicate the latter is your case. So IMHO, what you've done is OK. The clients can choose their means of managing Widget objects, as long as it doesn't affect other clients.
回答3:
You are overthinking this in my opinion. Everybody who wants to have a shared_ptr<Widget> is going to have to include the Widget header file anyway. Putting the typedef (which is a good idea imo) in Widget.h makes 100% sense to me.
回答4:
My approach (using underbar types, just because it is how I do it)
class Type
{
public:
typedef shared_ptr<Type> ptr;
typedef shared_ptr<const Type> const_ptr;
};
I have found the const_ptr version is pretty darn useful.
回答5:
I used to structure my C++ code into libraries. A library would have a bunch of headers for client consumption, all inside the include/LibraryName directory. Also, I would have one header called Fwd.h inside this directory with forward declarations of all classes along with their pointer typedefs.
In addition, each public header would include Fwd.h so that including the header would automatically give you all forward declarations and pointer typedefs. This worked really well in practice.
Not all classes are necessary to place in a shared_ptr though. I would only create pointer typedefs for types that I expected to be created dynamically, and in this case I would supply a factory. This has the added benefit that you may get away with supplying client code with interface types only, and hide concreted implementations in your library's src directory. Not specifically what you asked advice for, but this gives the complete picture of my method. As a final point, it's also possible to supply a convenience header called LibraryName.h that includes the Fwd.h and all other public headers.
Good luck!
回答6:
I generally use this approach to ease typing and makes a generic shared pointer interface for classes. Do note it's C++0x.
#include <iostream>
#include <memory>
template <class T>
struct SharedVirtual
{
typedef std::shared_ptr<T> VPtr;
};
template <class T>
struct Shared
{
typedef std::shared_ptr<T> Ptr;
template <class... P>
static Ptr ptr(P&&... args)
{
return std::make_shared<T>(std::forward<P>(args)...);
}
};
class Foo : public SharedVirtual<Foo>
{
public:
virtual void foo() const = 0;
};
class Test : public Foo, public Shared<Test>
{
public:
void foo() const { std::cout << "Hai u!" << std::endl; }
};
void print(const Foo::VPtr& ptr)
{
ptr->foo();
}
int main()
{
auto ptr = Test::ptr();
print(ptr);
}
回答7:
Second part first: use a namespace, i.e.:
namespace WidgetStuff {
class Widget { ..
typedef shared_ptr<Widget> WidgetPtr;
..
If you want to split it up:
namespace WidgetStuff {
class Widget { ...
}
...
namespace WidgetStuff {
typedef ...
You're the library author, you own the namespace, so no one else should invade it.
And now part one is answered too, if you choose you can do:
#include <widget.h>
#include <widget_utils.h>
by splitting up the namespace as above. The effect is no one has to use the utilities, whether or not they do they should not invade your namespace, so they're free to make WidgetPtr mean something else, as long as it isn't in your namespace.
来源：https://stackoverflow.com/questions/4429417/whats-the-best-strategy-for-typedefing-shared-pointers