我是 C 開發的新手。我正在嘗試運行彼此獨立的無限函式。問題陳述與此類似:

我試圖實作這一點的方式是
#include <iostream>
#include <cstdlib>
#include <pthread.h>
#include <unistd.h>
#include <mutex>
int g_i = 0;
std::mutex g_i_mutex; // protects g_i
// increment g_i by 1
void increment_itr()
{
const std::lock_guard<std::mutex> lock(g_i_mutex);
g_i = 1;
}
void *fun(void *s)
{
std::string str;
str = (char *)s;
std::cout << str << " start\n";
while (1)
{
std::cout << str << " " << g_i << "\n";
if(g_i > 1000) break;
increment_itr();
}
pthread_exit(NULL);
std::cout << str << " end\n";
}
void *checker(void *s) {
while (1) {
if(g_i > 1000) {
std::cout<<"**********************\n";
std::cout << "checker: g_i == 100\n";
std::cout<<"**********************\n";
pthread_exit(NULL);
}
}
}
int main()
{
int itr = 0;
pthread_t threads[3];
pthread_attr_t attr;
void *status;
// Initialize and set thread joinable
pthread_attr_init(&attr);
pthread_attr_setdetachstate(&attr, PTHREAD_CREATE_JOINABLE);
int rc1 = pthread_create(&threads[0], &attr, fun, (void *)&"foo");
int rc2 = pthread_create(&threads[1], &attr, fun, (void *)&"bar");
int rc3 = pthread_create(&threads[2], &attr, checker, (void *)&"checker");
if (rc1 || rc2 || rc3)
{
std::cout << "Error:unable to create thread," << rc1 << rc2 << rc3 << std::endl;
exit(-1);
}
pthread_attr_destroy(&attr);
std::cout << "main func continues\n";
for (int i = 0; i < 3; i )
{
rc1 = pthread_join(threads[i], &status);
if (rc1)
{
std::cout << "Error:unable to join," << rc1 << std::endl;
exit(-1);
}
std::cout << "Main: completed thread id :" << i;
std::cout << " exiting with status :" << status << std::endl;
}
std::cout << "main end\n";
return 0;
}
這可行,但我想知道此實作是否是執行此操作的標準方法,或者可以以更好的方式完成?
uj5u.com熱心網友回復:
您正確地在里面取了一個鎖increment_itr,但是您的fun函式在g_i沒有獲得鎖的情況下訪問。
改變這個:
void increment_itr()
{
const std::lock_guard<std::mutex> lock(g_i_mutex);
g_i = 1;
}
對此
int increment_itr()
{
std::lock_guard<std::mutex> lock(g_i_mutex); // the const wasn't actually needed
g_i = g_i 1;
return g_i; // return the updated value of g_i
}
這不是執行緒安全的:
if(g_i > 1000) break; // access g_i without acquiring the lock
increment_itr();
這更好:
if (increment_itr() > 1000) {
break;
}
需要類似的修復checker:
void *checker(void *s) {
while (1) {
int i;
{
std::lock_guard<std::mutex> lock(g_i_mutex);
i = g_i;
}
if(i > 1000) {
std::cout<<"**********************\n";
std::cout << "checker: g_i == 100\n";
std::cout<<"**********************\n";
break;
}
return NULL;
}
至于你的設計問題。這是根本問題。
You're proposing a dedicated thread that continuously takes a lock and would does some sort checking on a data structure. And if a certain condition is met, it would do some additional processing such as writing to a database. The thread spinning in an infinite loop would be wasteful if nothing in the data structure (the two maps) has changed. Instead, you only want your integrity check to run when something changes. You can use a condition variable to have the checker thread pause until something actually changes.
Here's a better design.
uint64_t g_data_version = 0;
std::conditional_variable g_cv;
void *fun(void *s)
{
while (true) {
<< wait for data from the source >>
{
std::lock_guard<std::mutex> lock(g_i_mutex);
// update the data in the map while under a lock
// e.g. g_n ;
//
// increment the data version to signal a new revision has been made
g_data_version = 1;
}
// notify the checker thread that something has changed
g_cv.notify_all();
}
}
Then your checker function only wakes up when it fun signals it to say something has changed.
void *checker(void *s) {
while (1) {
// lock the mutex
std::unique_lock<std::mutex> lock(g_i_mutex);
// do the data comparison check here
// now wait for the data version to change
uint64_t version = g_data_version;
while (version != g_data_version) { // check for spurious wake up
cv.wait(lock); // this atomically unlocks the mutex and waits for a notify() call on another thread to happen
}
}
}
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