如何在 DLL 中返回向量？

发布于 2024-10-23 19:18:31 字数 430 浏览 2 评论 0原文

我为我的应用程序开发了一个插件系统。我的软件是用 MFC/c++ 编写的，

我选择从 DLL 导入函数。但我面临一个问题。我已经使用 std 库进行了编写，但是 SO 向我表明“如果基于模板，则将任何内容传入和传出 DLL 时都会存在危险”，

所以我开始重写它。但现在我找不到好的解决方案。这个 dll 将解析 Youtube 并以如下结构返回结果：

struct VideoInfo{
wchar_t* strName;
wchar_t* strURL;
int iDuration;
float fRate;
};

但将是 50 个，在我之前的代码中，我只是返回了一个 std::vector。 *vecVideos; 但我怎样才能以安全的方式做到这一点？

原文

I developing a plugin system to my Application.
My software is written in MFC/c++

I Choose import the functions from a DLL.
But i facing a problem.
I had written using the std library, but SO showed my that "There is a danger when passing anything into and out of a DLL if it's based on a template"

So i started to rewrite it. But now i cannot found a good solution.
This dll will parse the Youtube and return the results in a struct like this:

struct VideoInfo{
wchar_t* strName;
wchar_t* strURL;
int iDuration;
float fRate;
};

But will be 50 of those, in my previous code i just returned a std::vector<VideoInfo*> *vecVideos; but how can i do that in a safe way ?

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尸血腥色 2024-10-30 19:18:31

您可以创建一个链接列表并返回它，而不是返回向量。

无需担心不兼容，而且使用起来很简单：

struct Node
{
    VideoInfo vi;
    Node *Next;
};

当您构建视频列表时，只需执行以下操作：

Node *ptr = new Node();
ptr->vi = MyVideoInfo; // Assumes MyVideoInfo is the new video
ptr->Next = current;
current = ptr; // Assuming current is our current linked list

您甚至可以在 c 中使用它，因为您只需将其包装在 extern "C " {} 并将 typedef 添加到节点声明的前面。

Instead of returning a vector, you can create a linked list and return that.

No need to worry about incompatibility, and it's simple to use:

struct Node
{
    VideoInfo vi;
    Node *Next;
};

When you're building your list of videos, just do the following:

Node *ptr = new Node();
ptr->vi = MyVideoInfo; // Assumes MyVideoInfo is the new video
ptr->Next = current;
current = ptr; // Assuming current is our current linked list

You could even use this in c, since you'd just need to wrap it in an extern "C" {} and add typedef to the front of your declaration of the node.

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