c++多線程編程：常見面試題

時間 2019-11-19

標籤 c++ 多線程編程常見面試欄目 C&C++ 简体版

原文原文鏈接

題目：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promise

子線程與主線程必有一個知足條件(flag == num),不知足條件的那個線程不可能獲取unique_lock(會在wait中釋放)，只有知足條件的線程才能獲取鎖，執行程序安全

mutex m;//保護條件的互斥訪問
condition_variable cond;//條件變量
int flag = 10;//條件
void fun(int num) {
    for (int i = 0; i<50; i++) {
        unique_lock<mutex> lk(m);//A unique lock is an object that manages a mutex object with unique ownership in both states: locked and unlocked.  
        while (flag != num)
            cond.wait(lk);//在調用wait時會執行lk.unlock()  
        for (int j = 0; j<num; j++)
            cout << j << " ";
        cout << endl;
        flag = (num == 10) ? 100 : 10;
        cond.notify_one();//被阻塞的線程喚醒後lk.lock()恢復在調用wait前的狀態  
    }
}
int main() {
    thread child(fun, 10);
    fun(100);
    child.join();
    
    system("pause");
    return 0;
}

題目：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，每一個線程將本身的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。多線程

mutex m;
condition_variable cond;
int loop = 10;
int flag = 0;

void func(int id)
{
    for (int i = 0; i < loop; ++i)
    {
        unique_lock<mutex> lk(m);
        while (flag != id)
            cond.wait(lk);
        cout << static_cast<char>('A' + id) << " ";
        flag = (flag + 1) % 3;
        cond.notify_all();
    }
}

void main()
{
    thread A(func, 0);
    thread B(func, 1);
    func(2);
    cout << endl;

    A.join();
    B.join();

    system("pause");
}

題目(google筆試題)：有四個線程一、二、三、4。線程1的功能就是輸出1，線程2的功能就是輸出2，以此類推.........如今有四個文件ABCD。初始都爲空。現要讓四個文件呈以下格式：
A：1 2 3 4 1 2....
B：2 3 4 1 2 3....
C：3 4 1 2 3 4....異步

D：4 1 2 3 4 1....oop

mutex m;
condition_variable cond;
int loop = 10;
int flag;

void func(int num)
{
    for (int i = 0; i < loop; ++i)
    {
        unique_lock<mutex> lk(m);
        while (num != flag)
            cond.wait(lk);
        cout << num + 1 << " ";
        flag = (flag + 1) % 4;
        cond.notify_all();
    }
}

void main(int argc,char *argv[])
{
    flag = atoi(argv[1]);
    thread one(func, 1);
    thread two(func, 2);
    thread three(func, 3);
    func(0);
    one.join();
    two.join();
    three.join();
    cout << endl;

    system("pause");
}

讀者寫者問題this

這也是一個很是經典的多線程題目，題目大意以下：有一個寫者不少讀者，多個讀者能夠同時讀文件，但寫者在寫文件時不容許有讀者在讀文件，一樣有讀者讀時寫者也不能寫。google

class rwlock {
private:
    mutex _lock;
    condition_variable _wcon, _rcon;
    unsigned _writer, _reader;
    int _active;
public:
    void read_lock() {
        unique_lock<mutex> lock(_lock);
        ++_reader;
        while (_active < 0 || _writer > 0)
            _rcon.wait(lock);
        --_reader;
        ++_active;
    }
    void write_lock() {
        unique_lock<mutex> lock(_lock);
        ++_writer;
        while (_active != 0)
            _wcon.wait(lock);
        --_writer;
        _active = -1;
    }
    void unlock() {
        unique_lock<mutex> lock(_lock);
        if (_active > 0) {
            --_active;
            if (_active == 0) _wcon.notify_one();
        }
        else {
            _active = 0;
            if (_writer > 0) _wcon.notify_one();
            else if (_reader > 0) _rcon.notify_all();
        }

    }
    rwlock() :_writer(0), _reader(0), _active(0) {
    }
};

void t1(rwlock* rwl) {
    while (1) {
        cout << "I want to write." << endl;
        rwl->write_lock();
        cout << "writing..." << endl;
        this_thread::sleep_for(chrono::seconds(5));
        rwl->unlock();
        this_thread::sleep_for(chrono::seconds(5));
    }
}

void t2(rwlock* rwl) {
    while (1) {
        cout << "t2-I want to read." << endl;
        rwl->read_lock();
        cout << "t2-reading..." << endl;
        this_thread::sleep_for(chrono::seconds(1));
        rwl->unlock();
    }
}

void t3(rwlock* rwl) {
    while (1) {
        cout << "t3-I want to read." << endl;
        rwl->read_lock();
        cout << "t3-reading..." << endl;
        this_thread::sleep_for(chrono::seconds(1));
        rwl->unlock();
    }
}

int main()
{
    rwlock* rwl = new rwlock();
    thread th1(t1, rwl);
    thread th2(t2, rwl);
    thread th3(t3, rwl);
    th1.join();
    th2.join();
    th3.join();

    system("pause");
    return 0;
}

線程安全的queueatom

STL中的queue是非線程安全的，一個組合操做：front(); pop()先讀取隊首元素而後刪除隊首元素，如果有多個線程執行這個組合操做的話，可能會發生執行序列交替執行，致使一些意想不到的行爲。所以須要從新設計線程安全的queue的接口。spa

template<typename T>
class threadsafe_queue
{
private:
    mutable std::mutex mut;
    std::queue<T> data_queue;
    std::condition_variable data_cond;
public:
    threadsafe_queue() {}
    threadsafe_queue(threadsafe_queue const& other)
    {
        std::lock_guard<std::mutex> lk(other.mut);
        data_queue = other.data_queue;
    }
    void push(T new_value)//入隊操做  
    {
        std::lock_guard<std::mutex> lk(mut);
        data_queue.push(new_value);
        data_cond.notify_one();
    }
    void wait_and_pop(T& value)//直到有元素能夠刪除爲止  
    {
        std::unique_lock<std::mutex> lk(mut);
        data_cond.wait(lk, [this] {return !data_queue.empty(); });
        value = data_queue.front();
        data_queue.pop();
    }
    std::shared_ptr<T> wait_and_pop()
    {
        std::unique_lock<std::mutex> lk(mut);
        data_cond.wait(lk, [this] {return !data_queue.empty(); });
        std::shared_ptr<T> res(std::make_shared<T>(data_queue.front()));
        data_queue.pop();
        return res;
    }
    bool try_pop(T& value)//無論有沒有隊首元素直接返回  
    {
        std::lock_guard<std::mutex> lk(mut);
        if (data_queue.empty())
            return false;
        value = data_queue.front();
        data_queue.pop();
        return true;
    }
    std::shared_ptr<T> try_pop()
    {
        std::lock_guard<std::mutex> lk(mut);
        if (data_queue.empty())
            return std::shared_ptr<T>();
        std::shared_ptr<T> res(std::make_shared<T>(data_queue.front()));
        data_queue.pop();
        return res;
    }
    bool empty() const
    {
        std::lock_guard<std::mutex> lk(mut);
        return data_queue.empty();
    }
};

題目：編寫程序完成以下功能：.net

1）有一int型全局變量g_Flag初始值爲0

2）在主線稱中起動線程1，打印「this is thread1」，並將g_Flag設置爲1

3）在主線稱中啓動線程2，打印「this is thread2」，並將g_Flag設置爲2

4）線程序1須要在線程2退出後才能退出

5）主線程在檢測到g_Flag從1變爲2，或者從2變爲1的時候退出

atomic<int> flag(0);

void worker1(future<int> fut) 
{//線程1  
    printf("this is thread1\n");
    flag = 1;
    fut.get();//線程1阻塞至線程2設置共享狀態  get等待異步操做結束並返回結果
    printf("thread1 exit\n");
}

void worker2(promise<int> prom) 
{//線程2  
    printf("this is thread2\n");//C++11的線程輸出cout沒有boost的好，仍是會出現亂序，因此採用printf，有點不爽  
    flag = 2;
    prom.set_value(10);//線程2設置了共享狀態後，線程1纔會被喚醒  
    printf("thread2 exit\n");
}

//利用promise future來控制線程退出的次序 int main()
{
    promise<int> prom;
    future<int> fut = prom.get_future();
    thread one(worker1, move(fut));//注意future和promise不容許拷貝，可是具有move語義  
    thread two(worker2, move(prom));
    while (flag.load() == 0);
　　///將本線程從調用線程中分離出來，容許本線程獨立執行
    one.detach();
    two.detach();
    //exit(1);//主線程到這裏退出  
    printf("main thread exit\n");

    system("pause");
    return 0;
}