在“if”内创建 COM 智能指针时出现错误 C2275陈述

发布于 2025-01-04 23:29:49 字数 480 浏览 3 评论 0原文

为什么我不能

if (IUnknownPtr p = anotherComPtr) {} //error C2275: 'IUnknownPtr' : illegal use of this type as an expression

，而我可以

if (int* a = anotherPointer) {}

IUnknownPtr 是通过 _COM_SMARTPTR_TYPEDEF(IUnknown, __uuidof(IUnknown)) 定义的（就像我使用的任何其他智能指针一样）

我如何创建一个 com smartptr 在 if 语句中并验证它是否有效？谢谢。

我用的是 VS

2008这与编码方式是否良好无关，而是与错误 C2275 相关。

原文

Why can't i

if (IUnknownPtr p = anotherComPtr) {} //error C2275: 'IUnknownPtr' : illegal use of this type as an expression

while i can

if (int* a = anotherPointer) {}

IUnknownPtr is defined throught _COM_SMARTPTR_TYPEDEF(IUnknown, __uuidof(IUnknown)) (like any others smart pointers i use)

How can i create a com smartptr within if statement and verify is it valid or not? Thank you.

I use VS 2008

p.s. This is not about is it good way of coding or not, it's about error C2275.

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悟红尘 2025-01-11 23:29:49

我无法在下面的小程序中重现您在 vs2008 中的编译器错误。您的包含文件、预处理器定义或编译器选项中可能存在不同的内容，从而导致不同的行为。

您可以在 if 语句之外声明一个 IUnknownPtr 类型的简单变量吗？

您可以使用下面的代码创建一个新项目而不出现错误吗？

以下任何一个都可以编译吗？

if (NULL == (IUnknownPtr ptr = someOtherPtr)) {
}

IUnknownPtr foo;
bool b(foo);

该错误表明编译器可以看到 IUnknownPtr 的定义，但无法将 IUnknownPtr 的赋值结果解释为 bool。

运算符 = 应该返回 IUnknownPtr& （已分配给的对象）。 _com_ptr_t 定义运算符 bool()。您的 _COM_SMARTPTR_TYPEDEF 是否生成对 _com_ptr_t 或其他类型的引用？您可以通过临时转储预处理器输出（属性/C++/预处理器/预处理到文件）来轻松找到答案

#include <comdef.h>

int main(int argc, char* argv[])
{
    IUnknownPtr foo;
    IUnknown* foo2 = NULL;

    if (IUnknownPtr foo3 = foo) {
        // do something
    }

    if (IUnknownPtr foo4 = foo2) {
        // do something
    }

    return 0;
}

I can't reproduce your compiler error in vs2008 in the small program below. There is likely to be something different in your include files, preprocessor definitions or compiler options which is giving you different behaviour.

Can you declare a simple variable of type IUnknownPtr outside an if statement?

Can you create a new project using the code below without the error?

Does either of the following compile OK?

if (NULL == (IUnknownPtr ptr = someOtherPtr)) {
}

IUnknownPtr foo;
bool b(foo);

The error suggests the compiler can see a definition of IUnknownPtr but can't interpret the result of the assignment of an IUnknownPtr as a bool.

operator = should return IUnknownPtr& (the object that's been assigned to). _com_ptr_t defines operator bool(). Are your _COM_SMARTPTR_TYPEDEF generating references to _com_ptr_t or some other type? You can easily find out by temporarily dumping out the preprocessor output (properties/C++/preprocessor/preprocess to a file)

#include <comdef.h>

int main(int argc, char* argv[])
{
    IUnknownPtr foo;
    IUnknown* foo2 = NULL;

    if (IUnknownPtr foo3 = foo) {
        // do something
    }

    if (IUnknownPtr foo4 = foo2) {
        // do something
    }

    return 0;
}

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