添加#pragma make_public（类型）不会删除C3767错误 [英] Adding #pragma make_public(Type) not removing C3767 error

问题描述

我有一个Assembly（A），它定义了一个Managed类，它有一个公共的构造函数，它接受两个本地类型。

I have an Assembly(A) which defines a Managed class which has a public constructor that takes two native types.

我可以访问Header文件和编译

I have access to the Header files and compiled lib files which contain the native types.

我创建了一个 C ++ / CLI 项目，并定义了一个 ref class ，它包含一个返回在（A）中定义的公共类型的单个 public：static 方法。

I created a C++/CLI project and defined a ref class which contains a single public: static method that returns a public type defined in (A).

当我试图通过传递一个原生类型，我收到了'C3767'YourType :: TypeB'：候选函数不可访问。

When I try to contstruct by passing in a native type I receive the `C3767 'YourType::TypeB': Candidate function(s) not accessible.

我添加了 #pragma make_public（Type）用于原生类型及其派生但仍然不快乐的任何类型。

I've added #pragma make_public(Type) for the native types and any type they derive from but still not joy.

我的类标题：

#pragma once
#include "StdAfx.h"

using namespace System;
using namespace AssemblyA;

namespace NativeWrapper {
    ref class MyFactory
    {
    public:
        static AssemblyAType^ Build();
    };
}

我的cpp文件：

#include "StdAfx.h"

#pragma make_public(nativeObjectRoot)
#pragma make_public(nativeObjectDerived)


#include "MyFactory.h"

using namespace System;

using namespace NativeWrapper;

AssemblyAType^ MyFactory::Build()
{
  nativeObjectDerived* myNativeObject;
  //myNativeObject initialised and set up here
  return gcnew AssemblyAType(myNativeObject); <--C3767
}

我看过和托管类型Assembly有一个带有这个签名的公共构造函数。似乎不能让编译指示工作？

I've looked and the managed type `AssemblyAType' has a public constructor with this signature. Can't seem to get the pragma to work??

所以总结一下。

我的C ++ / CLI项目引用定义在其构造函数中接受本机类型的类型的第三方程序集。我的项目也有头/ lib文件添加/链接到。

My C++/CLI project references a 3rd party Assembly that defines a type which takes a native type in its constructor. My project also has the header/lib files added/linked to.

注意：我上面的代码不是我已经得到，但我已经剥离相关部分。

Note: my code above isn't exactly what I've got but I've stripped out the pertinent parts.

推荐答案

make_public将使原生类型可见于使用它的程序集的使用者： http://msdn.microsoft.com/en-us/library/ ms235607（v = vs.80）.aspx 。它不会更改您引用的程序集中的可见性。

make_public will make the the native type visibile to consumers of the assembly where you use it: http://msdn.microsoft.com/en-us/library/ms235607(v=vs.80).aspx. It will not change the visibility in an assembly you reference.

似乎你引用的程序集应该有一个本地类型的make_public，或者简单地声明为本地类型public（见标题的类型可见性共享在本机和管理的客户端）。

It seems the assembly you referenced should have had either a make_public for the native type, or simply declared the native type public (see Type visibilty for a header Share a header file shared between native and managed clients).

下面的页面似乎表明他们应该得到一个警告，写入方法，而不使本机类型公开： http://msdn.microsoft.com/en-us/library /ms173713(v=vs.80).aspx

The below page seems to indicate that they should have gotten a warning for writing the method without making the native types public: http://msdn.microsoft.com/en-us/library/ms173713(v=vs.80).aspx

也许您可以发布第三方AssemblyCype代码，以确保没有其他的东西。

Perhaps you could post the third-party AssemblyAType code to make sure there's nothing else we're missing.

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