[C++ multithreading series] [2] The parameter transfer of the thread startup function requires type matching

When passing parameters to the execution function of the thread, the parameters must match, otherwise there will be compilation problems. Different compilers may handle it differently.

1.vs2013 compiled and passed.

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

typedef struct _A
{
	int a;
}A;

void f(A &a) 
{
	cout << a.a << endl;
	a.a++;
	cout << a.a << endl;
}



int main(int argc, int * argv[])
{
	A a = { 1 };

	cout << a.a << endl;

	// 这里vs2013编译通过，vs2017编译失败
	thread t(f,a);

	t.join();
	cout << a.a << endl;

	system("pause");
}

2.vs2017 compilation failed

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

typedef struct _A
{
	int a;
}A;

void f(A &a) 
{
	cout << a.a << endl;
	a.a++;
	cout << a.a << endl;
}



int main(int argc, int * argv[])
{
	A a = { 1 };

	cout << a.a << endl;

	// 这里vs2013编译通过，vs2017编译失败
	thread t(f,a);

	t.join();
	cout << a.a << endl;

	system("pause");
}

Prompt parameter type mismatch, that is, the required type of the function f parameter does not match the incoming parameter type.

The modification is as follows, and the compilation is passed.

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

typedef struct _A
{
	int a;
}A;

void f(A &a) 
{
	cout << a.a << endl;
	a.a++;
	cout << a.a << endl;
}



int main(int argc, int * argv[])
{
	A a = { 1 };

	cout << a.a << endl;

	thread t(f,std::ref(a));

	t.join();
	cout << a.a << endl;

	system("pause");
}

The result is as follows:

 

Summary: When passing parameters to the thread startup function, the types must match. Otherwise, it is not a compilation problem, or the running result is not as expected.