C++11 中的生产者-消费者

Question

我已经开始学习线程操作，并从一个旨在生成随机大写字母的简单程序开始。这些字母是随机生成的，并通过生产者添加到一个 char 数组中，任何添加的字母都以小写形式输出。消费者只需从 char 数组中输出常规大写字母。到目前为止，我有以下内容：

#include <iostream>           
#include <thread>             
#include <mutex>              
#include <condition_variable>
#include <random>

std::mutex mtx;
std::condition_variable cv;

int count = 0, buff_size = 0;
char* buff;

int random_int(int lower_bound) {
    std::random_device seed;
    std::mt19937 generator(seed());
    std::uniform_int_distribution<int> dist(lower_bound, std::nextafter(26, DBL_MAX));

    return dist(generator);
}

char random_char(int lower_bound) {
    return 'A' + (random_int(lower_bound) % 26);
}

/* Consumer

Reads characters from the buffer and prints them.

*/
void consume(int job) {
    std::unique_lock<std::mutex> lck(mtx);

    while (count == 0) {
        cv.wait(lck);
    }

    /*
    job + 1 = Running
    job = Ready
    */
    for (int i = 0; i < buff_size; i++) {
        std::cout << buff[i] << std::endl;
    }

    count--;
}

/* Producer

Randomly generates letters at (pos > buff_size & pos <= 26),
inserts them at the next available position in the buffer,
and then prints out the lowercase form of the inputted letter.

*/
void produce(int job) {
    std::unique_lock<std::mutex> lck(mtx);

    for (int i = 0; i < buff_size; i++) {
        buff[i] = random_char(buff_size);
        std::cout << tolower(buff[i]) << std::endl;
    }

    count++;
    cv.notify_one();
}

int main() {
    int buf_size = 0;

    std::cout << "The Producer-Consumer Problem (in C++11!)" << std::endl << "Enter the buffer size: ";
    std::cin >> buf_size;

    if (buf_size > 0 && buf_size <= 26) {
        // set the buffer size
        buff_size = buf_size;
        buff = new char[buff_size];
    }
    else {
        // rage quit
        exit(1);
    }

    std::thread production[10], processed[10];

    /* Initialize the arrays */
    for (int order = 0; order < buff_size; order++) {
        production[order] = std::thread(produce, order);
        processed[order] = std::thread(consume, order);
    }

    /* Join the threads to the main threads */
    for (int order = 0; order < buff_size; order++) {
        processed[order].join();
        production[order].join();
    }

    // free the allocated memory
    delete[] buff;
}

然而，我的输出是大写字母和随机数的混合体。怎么了？这是我第一次尝试，所以要温柔。 :)

Answer 1

output however is a mixture of capital letters and random numbers.

查看 zch 评论。我认为他的意思是：

而不是：

std::cout << tolower(buff[i]) << std::endl;

尝试

std::cout << char(tolower(buff[i])) << std::endl;

或

std::cout << (char)tolower(buff[i]) << std::endl;

因为 C++ 标签，你也许应该使用 static_cast

std::cout << static_cast<char>(tolower(buff[i])) << std::endl;

C++11 中的生产者-消费者

Producer-Consumer in C++11

c++

multithreading

consumer

producer

c++11