我有3个线程和2个共享资源,需要一些锁定......我试图用2个缓冲区来说明资源... - 线程1只能访问资源1 - 线程2可以访问资源1和2 - 线程3可以访问资源1和2
有人可以告诉我为什么以下锁定失败了吗?因为thread2和thread3将访问资源1和2 ...我想我可以使用try_lock? ......当thread2和thread3一次只能锁定1个互斥锁时,似乎弹出了问题......任何想法?
#include <iostream>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <vector>
#include <algorithm>
#include <cassert>
using namespace std;
class SynchronizationTest {
private:
mutex lock_r1;
mutex lock_r2;
vector<pair<string, int>> buffer_r1;
vector<pair<string, int>> buffer_r2;
unsigned int buffer_r1_max_size;
unsigned int buffer_r2_max_size;
bool buffer_r1_inc_element(const string &thread_id) {
if (buffer_r1.size() == buffer_r1_max_size) {
return true;
}
if (buffer_r1.size() == 0) {
buffer_r1.push_back(make_pair(thread_id, 0));
}
else {
int last_val = buffer_r1.back().second;
buffer_r1.push_back(make_pair(thread_id, ++last_val));
}
return false;
}
bool buffer_r2_inc_element(const string &thread_id) {
if (buffer_r2.size() == buffer_r2_max_size) {
return true;
}
if (buffer_r2.size() == 0) {
buffer_r2.push_back(make_pair(thread_id, 0));
}
else {
int last_val = buffer_r2.back().second;
buffer_r2.push_back(make_pair(thread_id, ++last_val));
}
return false;
}
public:
SynchronizationTest(int buff_r1_size, int buff_r2_size) : buffer_r1_max_size(buff_r1_size),
buffer_r2_max_size(buff_r2_size) {}
void thread1() {
bool buffer_r1_full = false;
while (!buffer_r1_full) {
{
unique_lock<mutex> l(lock_r1, std::defer_lock);
if (l.try_lock()) {
buffer_r1_full = buffer_r1_inc_element("thread1");
}
}
std::this_thread::sleep_for(std::chrono::milliseconds(10));
}
}
void thread2() {
bool buffer_r1_full = false;
bool buffer_r2_full = false;
while (!buffer_r1_full || !buffer_r2_full) {
{
unique_lock<mutex> lock1(lock_r1, defer_lock);
unique_lock<mutex> lock2(lock_r2, defer_lock);
int result = try_lock(lock1, lock2);
if(result == -1) {
buffer_r1_full = buffer_r1_inc_element("thread2");
buffer_r2_full = buffer_r2_inc_element("thread2");
}
else if(result != 0) {
buffer_r1_full = buffer_r1_inc_element("thread2");
}
else if(result != 1) {
buffer_r2_full = buffer_r2_inc_element("thread2");
}
}
std::this_thread::sleep_for(std::chrono::milliseconds(10));
}
}
void thread3() {
bool buffer_r1_full = false;
bool buffer_r2_full = false;
while (!buffer_r1_full || !buffer_r2_full) {
{
unique_lock<mutex> lock1(lock_r1, defer_lock);
unique_lock<mutex> lock2(lock_r2, defer_lock);
int result = try_lock(lock1, lock2);
if(result == -1) {
buffer_r1_full = buffer_r1_inc_element("thread3");
buffer_r2_full = buffer_r2_inc_element("thread3");
}
else if(result != 0) {
buffer_r1_full = buffer_r1_inc_element("thread3");
}
else if(result != 1) {
buffer_r2_full = buffer_r2_inc_element("thread3");
}
}
std::this_thread::sleep_for(std::chrono::milliseconds(10));
}
}
void print_buffer() {
for_each(buffer_r1.begin(), buffer_r1.end(), [](pair<string, int> p) { cout << p.first.c_str() << " " << p.second << endl; });
cout << '\n';
for_each(buffer_r2.begin(), buffer_r2.end(), [](pair<string, int> p) { cout << p.first.c_str() << " " << p.second << endl; });
}
};
int main() {
// your code goes here
SynchronizationTest st(20, 20);
thread t1(&SynchronizationTest::thread1, &st);
thread t2(&SynchronizationTest::thread2, &st);
thread t3(&SynchronizationTest::thread3, &st);
t1.join();
t2.join();
t3.join();
st.print_buffer();
return 0;
}
答案 0 :(得分:1)
std::try_lock
不起作用。如果返回-1,则保持所有锁定。如果它返回一个非负整数,没有锁定。返回的值告诉哪个锁失败,但是在try_lock
返回之前释放了成功锁定的所有锁。
答案 1 :(得分:0)
问题解决了:
unique_lock<mutex> lock1(lock_r1, defer_lock);
unique_lock<mutex> lock2(lock_r2, defer_lock);
bool result1 = lock1.try_lock();
bool result2 = lock2.try_lock();
if(result1 && result2) {
buffer_r1_full = buffer_r1_inc_element("thread2");
buffer_r2_full = buffer_r2_inc_element("thread2");
}
else if(result1) {
buffer_r1_full = buffer_r1_inc_element("thread2");
}
else if(result2) {
buffer_r2_full = buffer_r2_inc_element("thread2");
}