c++ - 通过实用程序 fn 将捕获式 lambda 传递给 C 样式回调 - 错误

Question

我想在实用函数的帮助下将捕获式 lambda 函数传递给 C 风格的回调:

#include <utility>
#include <iostream>
#include <array>

struct AWS_IoT_Client {};
struct IoT_Publish_Message_Params {
    int payload[1024];
};

typedef enum {
    SHADOW_ACK_TIMEOUT, SHADOW_ACK_REJECTED, SHADOW_ACK_ACCEPTED
} Shadow_Ack_Status_t;

typedef enum {
    SHADOW_GET, SHADOW_UPDATE, SHADOW_DELETE
} ShadowActions_t;

typedef void (*pApplicationHandler_t)(AWS_IoT_Client *pClient, char *pTopicName, uint16_t topicNameLen,
                                      IoT_Publish_Message_Params *pParams, void *pClientData);

typedef void (*fpActionCallback_t)(const char *pThingName, ShadowActions_t action, Shadow_Ack_Status_t status,
                                   const char *pReceivedJsonDocument, void *pContextData);

struct AWS {
    inline void subscribe(char*, pApplicationHandler_t, void*)
    {
    }

    inline void get_shadow(fpActionCallback_t, void *)
    {
    }
};

AWS* aws = new AWS;

    namespace utils {
      template<class F>
      struct c_style_callback_t {
        F f;
        template<class...Args>
        static void(*get_callback())(Args..., void*) {
          return [](Args...args, void* fptr)->void {
            (*static_cast<F*>(fptr))(std::forward<Args>(args)...);
          };
        }
        void* get_pvoid() {
          return std::addressof(f);
        }
      };
      template<class F>
      c_style_callback_t< std::decay_t<F> >
      c_style_callback( F&& f ) { return {std::forward<F>(f)}; }
    }

int main() {
    char someVar[1024]={0};

    auto task2 = utils::c_style_callback(
      [&] (const char *pThingName, ShadowActions_t action, Shadow_Ack_Status_t status,
           const char *pReceivedJsonDocument, void *pContextData) {
        //sprintf(someVar, "%s", text);
      }
    );

    aws->get_shadow(
      task2.get_callback<const char*, ShadowActions_t, Shadow_Ack_Status_t, const char*, void*>(),
      task2.get_pvoid()
    );

    auto task = utils::c_style_callback(
      [&] (AWS_IoT_Client *pClient, char *topicName, uint16_t topicNameLen, IoT_Publish_Message_Params *params) {
        char *text = (char *)params->payload;
        sprintf(someVar, "%s", text);
      }
    );

但是，我没有收到不同回调的错误:

    char topic[] = "some topic";
    aws->subscribe(
      topic,
      task.get_callback<AWS_IoT_Client*, char*, uint16_t, IoT_Publish_Message_Params*>(),
      task.get_pvoid()
    );
}

错误信息:

error: cannot initialize a parameter of type 'fpActionCallback_t' (aka 'void (*)(const char *, ShadowActions_t, Shadow_Ack_Status_t, const char *, void *)') with an rvalue of type 'void (*)(const char *, ShadowActions_t, Shadow_Ack_Status_t, const char *, void *, void *)': different number of parameters (5 vs 6)
      task2.get_callback<const char*, ShadowActions_t, Shadow_Ack_Status_t, const char*, void*>()`

我不明白为什么。

live example

更新:

我要强调的是，实用函数 c_style_callback 对以后遇到此问题的其他人非常有用。这是比答案 here 中的解决方案更灵活的解决方案因为这个支持自定义回调签名(不知道如何描述)。

我的问题更多是关于使用它而不是关于“将捕获作为函数指针的 lambda”的问题。

Answer 1

查看您的回调定义:

typedef void (*pApplicationHandler_t)(AWS_IoT_Client *pClient, char *pTopicName, uint16_t topicNameLen, IoT_Publish_Message_Params *pParams, void *pClientData);

typedef void (*fpActionCallback_t)(const char *pThingName, ShadowActions_t action, Shadow_Ack_Status_t status, const char *pReceivedJsonDocument, void *pContextData);

以及每种情况下的回调参数:

task.get_callback<AWS_IoT_Client*, char*, uint16_t, IoT_Publish_Message_Params*>()

task2.get_callback<const char*, ShadowActions_t, Shadow_Ack_Status_t, const char*, void*>()

在第一种情况下，您有一个附加参数 void* pClientData，在后一种情况下您没有，即使 get_callback 将返回一个带有该附加参数的 lambda ，因为您返回的是 return [](Args...args, void* fptr)，而不是 return [](Args...args)。

因此，要么将您的第二个 typedef 更改为:

typedef void (*fpActionCallback_t)(const char *pThingName, ShadowActions_t action, Shadow_Ack_Status_t status, const char *pReceivedJsonDocument, void *pContextData, void *pClientData);

或者改变get_callback的返回类型。