libcurl回调w / c ++类成员 [英] libcurl callbacks w/c++ class members

问题描述

取自libcurl网站上的 libcurl编程教程： / p>

Taken from the libcurl programming tutorial on the libcurl site:

libcurl与C ++

libcurl with C++

当接口libcurl时，C ++
而不是C：

There's basically only one thing to keep in mind when using C++ instead of C when interfacing libcurl:

回调不能是非静态类成员函数

The callbacks CANNOT be non-static class member functions

示例C ++代码：

class AClass 
{   
    static size_t write_data(void *ptr, size_t size, size_t nmemb,   void ourpointer)       
    {   /* do what you want with the data */   }  
}

这个限制是什么，是范围的问题吗？如果每个类成员都有自己的easy_handle，那么它然后可以使用回调它自己的非静态成员函数？它不会说。我可以用这种方式工作，但我很好奇为什么它存在。

What is this limitation for, is it a problem of scope? If each class member has it's own easy_handle then can it then use callbacks to it's own non static member functions? It doesn't go on to say. I can work with this either way, but I'm curious why it exists.

编辑：另一件事;这是否限制从类中创建easy_handle的？这将会对我的设计产生一个问题，如果是的话，我希望每个对象都有自己的easy_handle。

edit: One more thing; does this impose a limitation on creating easy_handle's from within a class? That would greatly pose a problem for my design if so, I'd like each object to have its own easy_handle.

推荐答案

非静态成员函数通常不允许作为C回调，因为它还需要实例指针作为隐藏参数，这是调用C代码不期望的（并且这个隐藏参数，取决于编译器，可能不会传递

A non-static member function is normally not allowed as a C callback because it neeeds also the instance pointer as a hidden parameter, which is not expected by the calling C code (and this hidden parameter, depending from the compiler, may not be passed on the stack, but on a particular register, ...).

静态成员函数通常实现为具有特定范围的普通函数与ABI无关）。

Static member functions, instead, are normally implemented as "normal" functions with a particular scope (which is not relevant as far as the ABI is concerned).

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