Today’s error series: How to use void * type function parameters

When I was working on a problem today, the void * type was used, so record it here

A rough description of the problem: how to use a function with void * type as a parameter

void Problem(void*  user_data)；

The following is an example of passing a shared pointer: the code is very simple, read it patiently.
Pay attention to two points: two long comments in the following code

#include <iostream>
using namespace std;
class StopMyDate {
    
    
public:
	string aaa;
	string bbb;
	string ccc;
};

void sss(void*  user_data)
{
    
    
    // 注意该处：类似于 *((int*)user_data)   /***************注意1*************/
	auto m_user_data = *((shared_ptr<StopMyDate>*)user_data);
	string aa = m_user_data->aaa;
	cout << "m_user_data->aaa : " << aa.c_str() << endl; //测试是否正确
}

int main()
{
    
    
	shared_ptr<StopMyDate>stopMyDate(new StopMyDate); // 定义一个共享指针,类型stopMyDate
	// 给成员赋值
	stopMyDate->aaa = "aaaaa";
	stopMyDate->bbb = "bbbbb";
	stopMyDate->ccc = "ccccc";
	
    // 此处调用，需注意添加(void*)  /***************注意2*************/
	sss((void *)&stopMyDate);
	return 0;
}

Output:

m_user_data->aaa : aaaaa

End:

Sharing is also a way to deepen your understanding of the problem again. It may not be comprehensive, but it is definitely useful and will continue to be improved later~