生産者と消費者の問題(C++実現)

コードの概要

TはT個の生産者がいることを示している.同時にT人の消費者もいる.そして各生産者は合計T個のデータを生産する.消費者1人当たり合計T個のデータを消費する.

Nは最大乱数の大きい値

を表す.

その信号量の実現、私は直接copyの授業の上でです.

コード#コード#
バージョン1

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

class Semaphore {
public:
    Semaphore(long count = 0) :count_(count) {}
    void Signal() {
        unique_lock lock(mutex_);
        ++count_;
        cv_.notify_one();
    }
    void Wait() {
        unique_lock lock(mutex_);
        //      false     ，
        //   Lambda         
        cv_.wait(lock, [=] { return count_ > 0; });
        --count_;
    }
    void setCount(long count = 0) {
        count_ = count;
    }
private:
    mutex mutex_;
    condition_variable cv_;
    long count_;
};


int T, N;
int BufferSize;
queue<string> output;
queue<int> Buffer;
Semaphore BufferChoose, BufferCnt, BufferLast;

void Poducer(int cnt) {
    srand(time(NULL));
    int data;
    for (int i = 0; i < T; ++i) {
        BufferLast.Wait();
        BufferChoose.Wait();
        data = rand() % N;
        Buffer.push(data);
        string temp = string("   ") + to_string(cnt) + string("       ") + to_string(data);
        output.push(temp);
        BufferChoose.Signal();
        BufferCnt.Signal();
    }
}

void Consumer(int cnt) {
    int data;
    for (int i = 0; i < T; ++i) {
        BufferCnt.Wait();
        BufferChoose.Wait();
        data = Buffer.front();
        Buffer.pop();
        string temp = string("   ") + to_string(cnt) + string("       ") + to_string(data) + string(",   ") + to_string(data * data);
        output.push(temp);
        BufferChoose.Signal();
        BufferLast.Signal();
    }
}

int main() {
    T = 4;
    N = 100;
    BufferSize = 2;
    _ASSERT(BufferSize > 0);

    BufferChoose.setCount(1);
    BufferCnt.setCount(0);
    BufferLast.setCount(BufferSize);

    vector vt;
    for (int i = 0; i < 2*T; ++i) {
        if (i < T) vt.push_back(thread(Poducer, i));
        else vt.push_back(thread(Consumer, i-T));
    }
    for (int i = 0; i < 2*T; ++i) {
        vt[i].join();
    }
    while (!output.empty()) {
        cout << output.front() << endl;
        output.pop();
    }
    system("pause");
}

バージョン2

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

class Semaphore {
public:
    Semaphore(long count = 0) :count_(count) {}
    void Signal() {
        unique_lock lock(mutex_);
        ++count_;
        cv_.notify_one();
    }
    void Wait() {
        unique_lock lock(mutex_);
        //      false     ，
        //   Lambda         
        cv_.wait(lock, [=] { return count_ > 0; });
        --count_;
    }
    void setCount(long count = 0) {
        count_ = count;
    }
private:
    mutex mutex_;
    condition_variable cv_;
    long count_;
};


int T, N;
int BufferSize;
queue<string> output;
queue<int> Buffer;
Semaphore BufferChoose, BufferCnt, BufferLast;

void Poducer(int cnt) {
    srand(time(NULL)+cnt);
    int data;
    for (int i = 0; i < T; ++i) {
        BufferLast.Wait();
        BufferChoose.Wait();
        data = rand() % N;
        Buffer.push(data);
        string temp = string("   ") + to_string(cnt) + string("       ") + to_string(data);
        output.push(temp);
        BufferChoose.Signal();
        BufferCnt.Signal();
    }
}

void Consumer(int cnt) {
    int data;
    for (int i = 0; i < T; ++i) {
        BufferCnt.Wait();
        BufferChoose.Wait();
        data = Buffer.front();
        Buffer.pop();
        string temp = string("   ") + to_string(cnt) + string("       ") + to_string(data) + string(",   ") + to_string(data * data);
        output.push(temp);
        BufferChoose.Signal();
        BufferLast.Signal();
    }
}

int main() {
    T = 4;
    N = 100;
    BufferSize = 10;
    _ASSERT(BufferSize > 0);

    BufferChoose.setCount(1);
    BufferCnt.setCount(0);
    BufferLast.setCount(BufferSize);

    vector vt;
    for (int i = 0; i < 2*T; ++i) {
        if (i < T) vt.push_back(thread(Poducer, i));
        else vt.push_back(thread(Consumer, i-T));
    }
    for (int i = 0; i < 2*T; ++i) {
        vt[i].join();
    }
    while (!output.empty()) {
        cout << output.front() << endl;
        output.pop();
    }
    system("pause");
}