检查 COM 指针是否相等

Check COM pointers for equality(检查 COM 指针是否相等)

问题描述

如果我有两个 COM 接口指针(即 ID3D11Texture2D)，并且我想检查它们是否是相同的底层类实例，我可以直接比较这两个指针是否相等?我已经看到在比较完成之前我们将其转换为其他内容的代码，所以想确认一下.

If I have two COM interface pointers (i.e. ID3D11Texture2D), and I want to check if they are the same underlying class instance, can I compare the two pointers directly for equality? I have seen code where we cast it to something else before the comparison is done, so wanted to confirm.

BOOL IsEqual (ID3D11Texture2D *pTexture1, ID3D11Texture2D *pTexture2)
{
    if (pTexture1 == pTexture2)
    {
        return true;
    }
    else
    {
        return false;
    }
} 

谢谢.

推荐答案

正确的 COM 方法是使用 IUnknown 查询接口.引用 这里 在 MSDN 中:

The correct COM way to do this is to query interface with IUnknown. A quote from the remarks here in MSDN:

对于任何一个对象，对任何一个对象上的 IUnknown 接口的特定查询对象的接口必须始终返回相同的指针值.这使客户端能够确定两个指针是否指向通过使用 IID_IUnknown 调用 QueryInterface 和相同的组件比较结果.具体而言，查询并非如此对于 IUnknown 以外的接口(即使相同的接口通过相同的指针)必须返回相同的指针值.

For any one object, a specific query for the IUnknown interface on any of the object's interfaces must always return the same pointer value. This enables a client to determine whether two pointers point to the same component by calling QueryInterface with IID_IUnknown and comparing the results. It is specifically not the case that queries for interfaces other than IUnknown (even the same interface through the same pointer) must return the same pointer value.

所以正确的代码是

BOOL IsEqual (ID3D11Texture2D *pTexture1, ID3D11Texture2D *pTexture2)
{
    IUnknown *u1, *u2;

    pTexture1->QueryInterface(IID_IUnknown, &u1);
    pTexture2->QueryInterface(IID_IUnknown, &u2);

    BOOL areSame = u1 == u2;
    u1->Release();
    u2->Release();

    return areSame;
}

更新

1. 添加了对 Release 的调用，以减少引用计数.感谢您的好评.
2. 您也可以使用 ComPtr 来完成这项工作.请查看 MSDN.

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