1、子線程循環 10 次,接着主線程循環 100 次,接着又回到子線程循環 10 次,接着再回到主線程又循環 100 次,如此循環50次,試寫出代碼
#include <iostream>
#include <thread>
#include <mutex>
#include <condition_variable>
std::mutex mtx;
std::condition_variable cv;
int flag = 10;
bool pred(int num)
{
return flag == num;
}
void fun(int num)
{
std::unique_lock<std::mutex> lck(mtx);
for (int i = 0; i < 3; i++)
{
while (flag != num)
cv.wait(lck);
for (int j = 0; j < num; j++)
std::cout << num << "sub fun " << j << std::endl;
flag = (num==10)?100:10;
cv.notify_one();
}
}
int main()
{
std::thread threads(fun, 10);
fun(100);
if (threads.joinable())
threads.join();
return 0;
}
2、編寫一個程序,開啓3個線程,這3個線程的ID分別爲A、B、C,每個線程將自己的ID在屏幕上打印10遍,要求輸出結果必須按ABC的順序顯示;如:ABCABC….依次遞推。
#include <thread>
#include <mutex>
#include <condition_variable>
#include <iostream>
std::mutex mtx;
std::condition_variable cv;
int flag = 0;
void fun(int num)
{
std::unique_lock<std::mutex> lck(mtx);
for (int i = 0; i < 10; i++)
{
while (flag != num)
cv.wait(lck);
std::cout << static_cast<char>('A' + num);
flag = (flag + 1) % 3;
cv.notify_all();//notify_one只能喚醒一個,並且不知道是哪個線程,若是兩個線程則可以用notify_one
}
}
int main()
{
std::thread th1(fun, 0);
std::thread th2(fun, 1);
std::thread th3(fun, 2);
if(th1.joinable())
th1.join();
if (th2.joinable())
th2.join();
if (th3.joinable())
th3.join();
return 0;
}