windows 如何从 CDC 对象获取 HDC 对象？
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原文地址: http://stackoverflow.com/questions/1234303/
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How can I get an HDC object from a CDC object?
提问by samoz
回答by Kirill V. Lyadvinsky
When you have CDCobject it will be implicitly converted to HDCwhen necessary:
当您拥有CDC对象时，它将HDC在必要时隐式转换为：
CDC dc;
HDC hdc = dc; // HDC hdc = dc.operator HDC();
If you have pointer to CDCobject then using function GetSafeHdcwill look more clear:
如果您有指向CDC对象的指针，则使用函数GetSafeHdc看起来会更清晰：
CDC* pdc = SOME;
HDC hdc = pdc->GetSafeHdc();
回答by Naveen
CDC class has operator HDC()defined which allows the compiler to convert a CDC object to HDC implicitly. Hence if you have CDC* and a function which takes HDC then you just dereference the pointer and send it to the function.
CDC 类已operator HDC()定义允许编译器将 CDC 对象隐式转换为 HDC。因此，如果您有 CDC* 和一个采用 HDC 的函数，那么您只需取消引用指针并将其发送给函数。
回答by CB Bailey
CDCis a C++ class which - to a reasonable approximation - encapsulates an HDC, which is a handle to a device context.
CDC是一个 C++ 类 - 合理地近似 - 封装了一个 HDC，它是一个设备上下文的句柄。
The documenation which you link to describes a conversion operator, which is a C++ construct that classes can supply to allow implicit conversion from an instance of a class to some other type. In this case the implicit conversion results in the underlying handle (HDC) which the CDCinstance encapsulates.
您链接到的文档描述了转换operator，这是一种 C++ 构造，类可以提供该构造以允许从类的实例到其他类型的隐式转换。在这种情况下，隐式转换会产生CDC实例封装的底层句柄 (HDC) 。
You can perform the conversion by using a CDCinstance anywhere were it needs to be converted to an HDC.
您可以在CDC任何需要将其转换为HDC.
Most simply:
最简单的：
void f( const CDC& cdc )
{
HDC hdc = cdc;
// use hdc here
}
回答by bdonlan
HDC hDC = dc;
回答by ChrisW
Just assign it.
只分配它。
CDC cdc = something.
HDC hdc = cdc;
if (hdc != 0)
{
//success...
}
