我正在尝试使用C ++ / WinRT在代码中实现自定义XAML控件。 但是，我尝试的实现无法编译。 作为概念的证明，我使用了这段代码：
#pragma once
#include <winrt/Windows.UI.Xaml.Controls.h>
namespace MyApp
{
struct MyControl : winrt::implements<MyControl, winrt::Windows::UI::Xaml::Controls::Control>
{
};
}
这导致以下编译器错误：
1>MyControl.cpp
1>c:\program files (x86)\windows kits\10\include\10.0.17134.0\cppwinrt\winrt\base.h(6416): error C2079: 'winrt::impl::producer<D,winrt::Windows::UI::Xaml::Controls::Control,void>::vtable' uses undefined struct 'winrt::impl::produce<D,I>'
1> with
1> [
1> D=MyApp::MyControl
1> ]
1>c:\program files (x86)\windows kits\10\include\10.0.17134.0\cppwinrt\winrt\base.h(7163): note: see reference to class template instantiation 'winrt::impl::producer<D,winrt::Windows::UI::Xaml::Controls::Control,void>' being compiled
1> with
1> [
1> D=MyApp::MyControl
1> ]
1>c:\xxx\mycontrol.h(8): note: see reference to class template instantiation 'winrt::implements<MyApp::MyControl,winrt::Windows::UI::Xaml::Controls::Control>' being compiled
我无法理解编译器错误。 显然，您无法像实现Windows运行时使用的其他类型一样实现XAML控件。
在代码中实现XAML自定义控件需要什么？
I am trying to implement a custom XAML control in code, using C++/WinRT. My attempted implementation, however, failed to compile. As a prove of concept I was using this code:
#pragma once
#include <winrt/Windows.UI.Xaml.Controls.h>
namespace MyApp
{
struct MyControl : winrt::implements<MyControl, winrt::Windows::UI::Xaml::Controls::Control>
{
};
}
This resulted in the following compiler error:
1>MyControl.cpp
1>c:\program files (x86)\windows kits\10\include\10.0.17134.0\cppwinrt\winrt\base.h(6416): error C2079: 'winrt::impl::producer<D,winrt::Windows::UI::Xaml::Controls::Control,void>::vtable' uses undefined struct 'winrt::impl::produce<D,I>'
1> with
1> [
1> D=MyApp::MyControl
1> ]
1>c:\program files (x86)\windows kits\10\include\10.0.17134.0\cppwinrt\winrt\base.h(7163): note: see reference to class template instantiation 'winrt::impl::producer<D,winrt::Windows::UI::Xaml::Controls::Control,void>' being compiled
1> with
1> [
1> D=MyApp::MyControl
1> ]
1>c:\xxx\mycontrol.h(8): note: see reference to class template instantiation 'winrt::implements<MyApp::MyControl,winrt::Windows::UI::Xaml::Controls::Control>' being compiled
I am unable to understand the compiler error. Apparently, you cannot implement a XAML control the same way you would implement other types for use by the Windows Runtime.
What is required to implement a XAML custom control in code?
最满意答案
WinRT中的“继承”或“子类化”与C ++继承略有不同。 因为这些是COM接口，所以当您继承WinRT运行时类时，您实际所做的是COM聚合 ，并结合实现基类型的可覆盖接口 。 由于COM聚合方面，这比标准C ++继承更加繁琐，所有委托/非委托，特殊构造等等。这将是WRL的一个主要痛苦，但C ++ / CX在把它抽象出去的引擎盖。 幸运的是，C ++ / WinRT可以帮助您提供两种类型的抽象，而不需要使用隐形魔法。
如果您正在创作一个不需要外部可见的类型（例如，应用程序，而不是运行时组件），C ++ / WinRT为此提供了方便的帮助：
#pragma once
#include <winrt/Windows.UI.Xaml.Controls.h>
namespace MyApp
{
struct MyControl : winrt::Windows::UI::Xaml::Controls::ControlT<MyControl>
{
void OnTapped(winrt::Windows::UI::Xaml::Input::TappedRoutedEventArgs const&);
};
}
此基类型ControlT将正确构造聚合的基本Control实例并将基本方法委托给它，同时还实现“可覆盖”接口。 这些可覆盖的方法都给出了一个占位符实现，默认调用基本方法，但您可以自己覆盖它们并获得自定义行为。
另一方面，如果您需要通过IDL创建一个预计的类型：
namespace MyApp
{
[default_interface]
runtimeclass MyControl : Windows.UI.Xaml.Controls.Control
{
MyControl();
};
}
这将产生与上面的内置ControlT案例类似的脚手架，但也会预测您的类型。 实际上，如果你检查这个类型的生成文件（在这个例子中，MyControl.gh），你会看到一个MyControlT ，它将全部连接起来。
（注意：只有当你有一个空的，可构造的，密封的运行时类时，才需要[default_interface]属性。一旦你添加成员，midl将会合成默认接口，而不需要任何其他的哄骗。
"Inheriting" or "subclassing" in WinRT is subtly different from C++ inheritance. Because these are COM interfaces, when you subclass a WinRT runtimeclass, what you're really doing is COM Aggregation, combined with implementing the base type's overridable interfaces. Due to the COM aggregation aspect, this is considerably more fussy than standard C++ inheritance, what with all the delegating/nondelegating, special construction, etc. This would be a major pain in WRL, but C++/CX did a bunch of compiler magic under the hood to abstract this away. Fortunately, C++/WinRT helps you out here with providing two types of abstractions, without resorting to invisible magic.
If you are authoring a type that doesn't need to be externally visible (e.g. an app, as opposed to a runtime component) C++/WinRT provides convenient helpers for this:
#pragma once
#include <winrt/Windows.UI.Xaml.Controls.h>
namespace MyApp
{
struct MyControl : winrt::Windows::UI::Xaml::Controls::ControlT<MyControl>
{
void OnTapped(winrt::Windows::UI::Xaml::Input::TappedRoutedEventArgs const&);
};
}
This base type ControlT will correctly construct the aggregated base Control instance and delegate base methods to it, while also implementing the "overridable" interfaces. These overridable methods are all given a placeholder implementation that defaults to calling the base method, but you can override them yourself and get your custom behavior.
If, on the other hand, you need to author a type that's projected, via IDL:
namespace MyApp
{
[default_interface]
runtimeclass MyControl : Windows.UI.Xaml.Controls.Control
{
MyControl();
};
}
That will generate similar scaffolding as the built-in ControlT case above, but also projects your type. In fact, if you examine the generated file for this type (in this example, MyControl.g.h), you'd see a MyControlT where that gets all hooked up.
(Note: the [default_interface] attribute is only needed if you have an empty, constructible, sealed runtimeclass. Once you add members, midl will figure synthesize the default interface without any other coaxing.
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