【发布时间】:2020-06-09 03:13:32
【问题描述】:
我找不到任何错误,程序没有显示任何输出;但是,如果我在看到一些输出之后在 number() 中引入一些 print 的 cout,是否会丢失信号?
我已经启动了 4 个线程,所有线程都在一个公共互斥锁和一个条件变量上使用了 unique_lock。如果我使用信号量,则相同的逻辑可以工作,但是当我将它们转换为 c++11 condition_variable 时,我在屏幕上看不到任何输出。线程挂起并等待信号。但是线程[3] 应该运行,因为它等待的条件为真;当前=1。
#include<iostream>
#include<thread>
#include<mutex>
using namespace std;
class FizzBuzz {
private:
int n;
int curr;
std::mutex mtx;
std::condition_variable cv;
public:
FizzBuzz(int n) {
this->n = n;
curr = 1;
}
// printFizz() outputs "fizz".
void fizz(function<void()> printFizz) {
while(curr <= n) {
std::unique_lock<std::mutex> lck(mtx);
cv.wait(lck, [&]{return (curr <= n && curr % 3 == 0 && curr % 5 != 0);});
printFizz();
curr++;
lck.unlock();
cv.notify_all();
}
}
// printBuzz() outputs "buzz".
void buzz(function<void()> printBuzz) {
while(curr <= n) {
std::unique_lock<std::mutex> lck(mtx);
cv.wait(lck, [&]{return (curr <= n && curr % 3 != 0 && curr % 5 == 0);});
printBuzz();
curr++;
lck.unlock();
cv.notify_all();
}
}
// printFizzBuzz() outputs "fizzbuzz".
void fizzbuzz(function<void()> printFizzBuzz) {
while(curr <= n) {
std::unique_lock<std::mutex> lck(mtx);
cv.wait(lck, [&]{return (curr <= n && curr % 3 == 0 && curr % 5 == 0);});
printFizzBuzz();
curr++;
lck.unlock();
cv.notify_all();
}
}
// printNumber(x) outputs "x", where x is an integer.
void number(function<void(int)> printNumber) {
while(curr <= n) {
std::unique_lock<std::mutex> lck(mtx);
cv.wait(lck, [&]{return (curr <= n && curr % 3 != 0 && curr % 5 != 0);});
printNumber(curr);
curr++;
lck.unlock();
cv.notify_all();
}
}
};
void printFizz (void) {
cout << "Fizz";
};
void printBuzz (void) {
cout << "Buzz";
};
void printFizzBuzz (void) {
cout << "FizzBuzz";
};
void printNumber (int n) {
cout << n;
};
int main(int argc, char* argv[]) {
int num;
sscanf(argv[1], "%d", &num);
std::function<void(void)> f_fizz = printFizz;
std::function<void(void)> f_buzz = printBuzz;
std::function<void(void)> f_fizzbuzz = printFizzBuzz;
std::function<void(int)> f_num = printNumber;
FizzBuzz *objPtr = new FizzBuzz(num);
std::thread threads[4];
threads[0] = std::thread(&FizzBuzz::fizz, objPtr ,f_fizz);
threads[1] = std::thread(&FizzBuzz::buzz,objPtr, f_buzz);
threads[2] = std::thread(&FizzBuzz::fizzbuzz,objPtr, f_fizzbuzz);
threads[3] = std::thread(&FizzBuzz::number,objPtr, f_num);
for (auto& th: threads) th.join();
return 0;
}
【问题讨论】:
标签: multithreading c++11 concurrency condition-variable fizzbuzz