Question

在Visual Studio 2017中，如果我创建一个基本的C＃库并将其导出为DLL，然后创建一个TLB，我可以在C ++项目中#import TLB，它似乎可以成功运行。但是，GetRecordInfoFromGuids（）始终因E_INVALIDARG而失败。我认为问题是生成的TLH文件在一个地方缺少UUID。这就是我所拥有的：

Base.cs

using System.Windows.Forms;
Namespace Test {
    public class Base : IBase {
        public void BaseFoo() {
            MessageBox.Show("Test");
        }
    }
}

IBase.cs

Namespace Test {
    public inteface IBase {
       void BaseFoo();
    }
}

TestBase.cpp

#import "path\to\Test.tlb"
void TestBase() {
    HRESULT hrCoInit = 0;
    hrCoInit = CoInitialize(NULL);

    if (!SUCCEEDED(hrCoInit)) { return; }

    IRecordInfoPtr recordInfoBase = NULL;
    HRESULT hrRIBase = GetRecordInfoFromGuids(
        Test::LIBID_Test, 
        1, 
        0, 
        0, 
        Test::CLSID_Base, 
        &recordInfoBase
    ); /* always results in E_INVALIDARG */
}

test.tlh

// Created by Microsoft (R) C/C++ Compiler Version 14.11.25547.0 (8f5b6942).
//
// d:\garre\documents\source\test\testlib\debug\test.tlh
//
// C++ source equivalent of Win32 type library ..\Test\bin\Debug\Test.tlb
// compiler-generated file created 01/29/18 at 22:51:17 - DO NOT EDIT!

#pragma once
#pragma pack(push, 8)

#include <comdef.h>

namespace Test {

//
// Forward references and typedefs
//

struct __declspec(uuid("066dd02f-1f6d-460c-8833-d22344da22e6"))
/* LIBID */ __Test;
/* ************************** Missing UUID Here? ***********************/
struct /* coclass */ Base;
struct __declspec(uuid("d123a47b-b549-3e35-baa6-63fbf1d2e996"))
/* dual interface */ IBase;
struct __declspec(uuid("49a2c4f5-cb61-3192-a3c5-fb90c6bab315"))
/* dual interface */ _Base;

//
// Smart pointer typedef declarations
//

_COM_SMARTPTR_TYPEDEF(IBase, __uuidof(IBase));
_COM_SMARTPTR_TYPEDEF(_Base, __uuidof(_Base));

//
// Type library items
//

struct __declspec(uuid("7b25bdc6-43c0-34db-ac1d-5f4e6a791982"))
Base;
    // [ default ] interface _Base
    // interface _Object
    // interface IBase

struct __declspec(uuid("d123a47b-b549-3e35-baa6-63fbf1d2e996"))
IBase : IDispatch
{
    //
    // Raw methods provided by interface
    //

      virtual HRESULT __stdcall BaseFoo ( ) = 0;
};

struct __declspec(uuid("49a2c4f5-cb61-3192-a3c5-fb90c6bab315"))
_Base : IDispatch
{};

//
// Named GUID constants initializations
//

extern "C" const GUID __declspec(selectany) LIBID_Test =
    {0x066dd02f,0x1f6d,0x460c,{0x88,0x33,0xd2,0x23,0x44,0xda,0x22,0xe6}};
extern "C" const GUID __declspec(selectany) CLSID_Base =
    {0x7b25bdc6,0x43c0,0x34db,{0xac,0x1d,0x5f,0x4e,0x6a,0x79,0x19,0x82}};
extern "C" const GUID __declspec(selectany) IID_IBase =
    {0xd123a47b,0xb549,0x3e35,{0xba,0xa6,0x63,0xfb,0xf1,0xd2,0xe9,0x96}};
extern "C" const GUID __declspec(selectany) IID__Base =
    {0x49a2c4f5,0xcb61,0x3192,{0xa3,0xc5,0xfb,0x90,0xc6,0xba,0xb3,0x15}};

} // namespace Test

#pragma pack(pop)

我尝试使用__uuidof（Test :: Base），使用不同的LCID，覆盖Base.cs中的UUID类Base [ComVisible（true）]，但这些似乎都没有任何效果。我可以使用Test :: BasePtr创建一个实例，所以其他一切似乎都是正确的;我只是无法得到任何引用IRecordInfo的东西。

在设置中设置“使程序集COM-Visible”。我提供了签名的钥匙。使用RegAsm.exe或使用“注册COM互操作”会产生同样的结果。

导入的TLB导致TLH中缺少信息

0 个答案: