下面的生产者/消费者程序应该一次将一个字符传送到缓冲区，然后打印它。该程序最初运行，但随后总是在使用者循环的第三次迭代中失败。 -pthread包含在编译时。
程序应该能够遍历'poem'变量，最终导致逐个字符地打印出整首诗。
#include <iostream>
#include <cmath>
#include <cstdlib>
#include <thread>
#include <fstream>
#include <chrono>
#include <mutex>
#include <condition_variable>
#include <vector>
#include <queue>
using namespace std;
mutex mtx;
condition_variable cond_var;
int thread_ID = 1;
vector<char> txtImport(vector<char> output){
ifstream file("poem.txt");
char character;
while (file.get(character)){
output.push_back(character);
}
return output;
}
void producer(vector<char> poem, char* buffer, int poem_size) {
cout << "\n\nPROD- \n";
int poem_position = 68;
const int producer_ID = 1;
unique_lock<mutex> lock(mtx);
while(true){
cout << "\n1";
if(poem_position == poem_size) {exit(1);}
if(thread_ID!=producer_ID) {cond_var.wait(lock);}
else{
cout << "\nlock\n";
*buffer = poem[poem_position];
poem_position += 1;
thread_ID = 2;
cout << poem_position << " 2 \n";
cout << poem[poem_position] << " 4 \n";
cout << "CHAR: " << *buffer << " \n------\n\n";
lock.unlock();
cond_var.notify_all();
std::this_thread::sleep_for(std::chrono::milliseconds(1000));
}
}
}
void consumer(char* buffer){
cout << "\n\n -CONS\n";
const int consumer_ID = 2;
unique_lock<mutex> lock(mtx);
while(true){
cout << "\n one ";
if(thread_ID!=consumer_ID) {cond_var.wait(lock);}
else{
cout << "\n lock\n";
cout << " ->" << *buffer << " <-\n\n";
thread_ID = 1;
lock.unlock();
cond_var.notify_all();
std::this_thread::sleep_for(std::chrono::milliseconds(1000));
}
}
}
int main(void) {
vector<char> poem;
poem = txtImport(poem);
char buffer = 'z';
int poem_size = poem.size();
cout << "\n--MAIN--\n";
thread thread_two(consumer, &buffer);
thread thread_one(producer, poem, &buffer, poem_size);
thread_one.join();
thread_two.join();
return 0;
}
程序输出为：
--MAIN--
-CONS
one
PROD-
1
lock
69 2
b 4
CHAR: m
------
one
lock
->m <-
1
lock
70 2
e 4
CHAR: b
------
one
lock
->b <-
terminate called after throwing an instance of 'std::system_error'
terminate called recursively
what(): Aborted (core dumped)
您有不确定的行为。
cond_var.wait被调用时，传递的互斥锁必须被当前线程锁定。它不会在您的代码中发生（第一次迭代除外）。
您的情况：
unique_lock<mutex> lock(mtx);
while (true)
cond_var.wait(lock);
lock.unlock();
所以在while循环的第二次迭代中，当wait被调用时，lock被解锁。错误。
要修复，将unique_lock构造移到both]中（消费者/生产者函数中的相同问题）while循环。
lock-类型为std :: unique_lock的对象，必须为被当前线程锁定