C ++ class object definitions Considerations

#include <iostream>
using namespace std;

class background_task
{
public:
	background_task()
	{
		cout << "默认构造" << this->m_num << endl;
	}
	background_task(int num)
	{
		this->m_num = num;
		cout << "带参构造" << this->m_num << endl;
	}
	//void operator()()
	//{
	//	cout << "background_task.....operator..." << endl;
	//}

private:
	int m_num = 20;
};

int main()
{
	background_task b1;		//1
	background_task b2();	//2
	background_task b3(10); //3

	// 无区别
	background_task *p1 = new background_task;
	background_task *p2 = new background_task();

	//对于内置类型而言,加括号是进行了初始化，不加是未进行初始化
	int *p1Int = new int();
	int *p2Int = new int;


	//b1();

	background_task();	  //匿名对象
	background_task(199); //匿名对象
	background_task bts1 = background_task();	// 同background_task b1;
	background_task bts2 = background_task(10);// 同background_task b3(10);;
	return 0;
}

Analysis declare variables defined above three kinds of differences:

1, define local variables b1, automatically call the constructor function parameters without a background_task.

2, Special Note: here is equivalent to declare a function return value of the function name is background_task b2 just a statement, not implemented.

3, the definition of local variables b3, parameterized constructor automatically calls the class background_task

Note:

Comments in the code section requires special attention ::

In this case B1 () function is called overloading function void operator () ()

 

Sometimes we use an anonymous object passed to the function parameters, but sometimes differs from the expected:

#include <iostream>
#include <thread>
using namespace std;

class background_task
{
public:
	background_task()
	{
		cout << "默认构造" << this->m_num << endl;
	}
	background_task(int num)
	{
		this->m_num = num;
		cout << "带参构造" << this->m_num << endl;
	}

	void operator()()
	{

		cout << "background_task...Threadid:" << this_thread::get_id() << endl;
		cout << "Threadnum:" << this->m_num << endl;
	}

private:
	int m_num = 20;
};

int main()
{
	//那么此时编译器会把background_task()解释为函数声明background_task (__cdecl *)(void)
	//std::thread   t(background_task()); 

	// 解决上述问题两种方案
	std::thread   t1((background_task()));
	std::thread   t2{background_task(100)};

	int numA = 2000;
	std::thread  t3([&](){
		cout << "lambda" << numA << endl;
		numA = 1000;
	});

	
	t1.join();
	t2.join();
	t3.join();
	cout << "main numA=" << numA;
	return 0;
}