从extern C struct继承时的成员初始化

In Mixing C and C++ Code in the Same Program the following example (slightly abbreviated here to the relevant parts) is given. Assume buf.h contains the following:

在将C和c++代码混合到同一个程序中时，给出了下面的示例(此处对相关部分略作简化)。假设缓冲区。h包含以下:

struct buf {
    char* data;
    unsigned count;
};

// some declarations of existing C functions for handling buf...

It is then recommended to use

然后建议使用它

extern "C" {
  #include "buf.h"
}

class mybuf : public buf {
public:
    mybuf() : data(0), count(0) { }

    // add new methods here (e.g. wrappers for existing C functions)...
};

in order to use the struct within C++ with added features.

为了在c++中使用带有附加特性的结构体。

However, this clearly will produce the following error:

但是，这显然会产生以下错误:

error: class `mybuf' does not have any field named `data'
error: class `mybuf' does not have any field named `count'

The reasons for this are explained in How can I initialize base class member variables in derived class constructor?, C++: Initialization of inherited field, and Initialize parent's protected members with initialization list (C++).

原因是如何在派生类构造函数中初始化基类成员变量?， c++:初始化继承字段，用初始化列表(c++)初始化父级受保护成员。

Thus, I have the following two questions:

因此，我有以下两个问题:

Is the code provided just plainly wrong or am I missing some relevant aspect? (After all, the article seems to stem from a reputable source)
代码是否提供了明显的错误，或者我是否遗漏了一些相关的方面?(毕竟，这篇文章的出处似乎很好)
What is the correct way to achieve the desired effect (i.e., turning a C struct into a C++ class and adding some convenience methods like, e.g., a constructor, etc.)?
达到预期效果的正确方法是什么?，将C struct转换为c++类，并添加一些方便的方法，例如，构造函数等)?

Update: Using aggregation initialization as suggested, i.e.,

更新:按照建议使用聚合初始化，即，

mybuf() : buf{0, 0} {}

works, but requires C++11. I therefore add the following question:

工作,但是需要c++ 11。因此，我增加了以下问题:

Using C++03, is there a better way to achieve the desired outcome than using the following constructor?

使用c++ 03，是否有比使用以下构造函数更好的方法来实现所需的结果?
```
mybuf() {
  data = 0;
  count = 0;
}
```

4 个解决方案

#1

If you can use a c++11 compatible compiler then this would be a perfect use case for an initializer list using aggregate initialization.

如果您可以使用与c++11兼容的编译器，那么对于使用聚合初始化的初始化器列表来说，这将是一个完美的用例。

mybuf() : buf{0, 0}
{}

#2

One "correct" way, if your compiler is C++11 capable, is to use e.g.

一个“正确”的方法，如果你的编译器是c++ 11，它是用例。

mybuf() : buf{0, 0} {}

#3

This has nothing to do with mixing C and C++. You're trying to initialise members that don't exist; that they exist in a base class isn't enough. You need to initialise the base itself.

这与混合C和c++没有任何关系。你试图初始化不存在的成员;它们存在于基类中是不够的。你需要初始化基地本身。

In this case, use aggregate initialisation:

在这种情况下，使用聚合初始化:

class mybuf : public buf
{
public:
    mybuf() : buf{0, 0} {}
};

#4

class mybuf : public buf {
public:
    mybuf();    
    // add new methods here (e.g. wrappers for existing C functions)...
};

const buf init = {0,0};

mybuf::mybuf() : buf(init) {};

will work.

将工作。

I have seen this work with some compilers, but don't have a standard handy to check if it is standard or an extension.

我已经看到了一些编译器的工作，但是没有一个标准来检查它是标准的还是扩展的。

class mybuf : public buf {
public:
    mybuf() : buf(init) { }

    // add new methods here (e.g. wrappers for existing C functions)...

    private:

    const buf init = {0,0};
};

#1