Does passing parameters in lambda's capture to boost asio post/dispatch thread safe?
I'm using lambda's capture in order to pass parameters to boost::asio::io_context::post callback.
Is it thread safe?
Code
#include <iostream>
#include "boost/asio.hpp"
#include <thread>
int main() {
boost::asio::io_service io_service;
boost::asio::io_service::work work(io_service);
std::thread t([&](){
io_service.run();
});
auto var = 1;
io_service.post([&io_service, var]() {
std::cout << "v: " << var << std::endl;
io_service.stop();
});
t.join();
return 0;
}
As you can see, I pass var in the lambda's capture.
the main thread sets var's value, and thread t reads it.
I didn't use any of memory ordering, for example, std::memory_order_release after setting var to 1, and std::memory_order_acquire before reading var value. Even more, I don't think I can - because the variable var is passed by value to the lambda.
Is it safe to do that?
If not, how should it be done?
multithreading c++11 boost synchronization boost-asio
asked Nov 12 at 19:33
hudac
96621240
add a comment |
I'm using lambda's capture in order to pass parameters to boost::asio::io_context::post callback.
Is it thread safe?
Code
#include <iostream>
#include "boost/asio.hpp"
#include <thread>
int main() {
boost::asio::io_service io_service;
boost::asio::io_service::work work(io_service);
std::thread t([&](){
io_service.run();
});
auto var = 1;
io_service.post([&io_service, var]() {
std::cout << "v: " << var << std::endl;
io_service.stop();
});
t.join();
return 0;
}
As you can see, I pass var in the lambda's capture.
the main thread sets var's value, and thread t reads it.
I didn't use any of memory ordering, for example, std::memory_order_release after setting var to 1, and std::memory_order_acquire before reading var value. Even more, I don't think I can - because the variable var is passed by value to the lambda.
Is it safe to do that?
If not, how should it be done?
multithreading c++11 boost synchronization boost-asio
asked Nov 12 at 19:33
hudac
96621240
add a comment |
I'm using lambda's capture in order to pass parameters to boost::asio::io_context::post callback.
Is it thread safe?
Code
#include <iostream>
#include "boost/asio.hpp"
#include <thread>
int main() {
boost::asio::io_service io_service;
boost::asio::io_service::work work(io_service);
std::thread t([&](){
io_service.run();
});
auto var = 1;
io_service.post([&io_service, var]() {
std::cout << "v: " << var << std::endl;
io_service.stop();
});
t.join();
return 0;
}
As you can see, I pass var in the lambda's capture.
the main thread sets var's value, and thread t reads it.
I didn't use any of memory ordering, for example, std::memory_order_release after setting var to 1, and std::memory_order_acquire before reading var value. Even more, I don't think I can - because the variable var is passed by value to the lambda.
Is it safe to do that?
If not, how should it be done?
multithreading c++11 boost synchronization boost-asio
asked Nov 12 at 19:33
hudac
96621240
I'm using lambda's capture in order to pass parameters to boost::asio::io_context::post callback.
Is it thread safe?
Code
#include <iostream>
#include "boost/asio.hpp"
#include <thread>
int main() {
boost::asio::io_service io_service;
boost::asio::io_service::work work(io_service);
std::thread t([&](){
io_service.run();
});
auto var = 1;
io_service.post([&io_service, var]() {
std::cout << "v: " << var << std::endl;
io_service.stop();
});
t.join();
return 0;
}
As you can see, I pass var in the lambda's capture.
the main thread sets var's value, and thread t reads it.
I didn't use any of memory ordering, for example, std::memory_order_release after setting var to 1, and std::memory_order_acquire before reading var value. Even more, I don't think I can - because the variable var is passed by value to the lambda.
Is it safe to do that?
If not, how should it be done?
multithreading c++11 boost synchronization boost-asio
multithreading c++11 boost synchronization boost-asio
asked Nov 12 at 19:33
hudac
96621240
asked Nov 12 at 19:33
hudac
96621240
asked Nov 12 at 19:33
hudac
96621240
asked Nov 12 at 19:33
hudac
96621240
asked Nov 12 at 19:33
hudac
96621240
96621240
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
It is thread-safe.
Closure object is created by main thread (with copying var value) after var was created and initialized.
Next, closure object is passed as argument to post method which queues this function object and returns immediately without calling functor. Functor is called between post and t.join calls - post guarantees it.
So your code must be thread-safe.
You would need some synchronization method (for example, use of mutex+lock_guard)
if var was passed by reference [1] and some writing operations on var [2]
were performed between post and t.join calls:
auto var = 1;
io_service.post([&io_service, &var]() { // [1] takes by reference
std::cout << "v: " << var << std::endl; // lock mutex for printing
io_service.stop();
});
var = 10; // [2] , lock mutex for writing
// synchronization must be added because between post and t.join calls,
// reading and writing operations are executed
t.join();
in this case you would have to protect var.
answered Nov 12 at 21:49
rafix07
6,6131613
So, are you saying: as thelambdapasses into threadtand get queued and executed successfully, the variable also passes?
– hudac
Nov 13 at 7:41
Ifvaris passed by value to lambda, closure object holds it as data member (copy of thisvar), so while passing closure to thread , we can say that the copy ofvaris passed.
– rafix07
Nov 13 at 7:56
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
It is thread-safe.
Closure object is created by main thread (with copying var value) after var was created and initialized.
Next, closure object is passed as argument to post method which queues this function object and returns immediately without calling functor. Functor is called between post and t.join calls - post guarantees it.
So your code must be thread-safe.
You would need some synchronization method (for example, use of mutex+lock_guard)
if var was passed by reference [1] and some writing operations on var [2]
were performed between post and t.join calls:
auto var = 1;
io_service.post([&io_service, &var]() { // [1] takes by reference
std::cout << "v: " << var << std::endl; // lock mutex for printing
io_service.stop();
});
var = 10; // [2] , lock mutex for writing
// synchronization must be added because between post and t.join calls,
// reading and writing operations are executed
t.join();
in this case you would have to protect var.
answered Nov 12 at 21:49
rafix07
6,6131613
So, are you saying: as thelambdapasses into threadtand get queued and executed successfully, the variable also passes?
– hudac
Nov 13 at 7:41
Ifvaris passed by value to lambda, closure object holds it as data member (copy of thisvar), so while passing closure to thread , we can say that the copy ofvaris passed.
– rafix07
Nov 13 at 7:56
add a comment |
It is thread-safe.
Closure object is created by main thread (with copying var value) after var was created and initialized.
Next, closure object is passed as argument to post method which queues this function object and returns immediately without calling functor. Functor is called between post and t.join calls - post guarantees it.
So your code must be thread-safe.
You would need some synchronization method (for example, use of mutex+lock_guard)
if var was passed by reference [1] and some writing operations on var [2]
were performed between post and t.join calls:
auto var = 1;
io_service.post([&io_service, &var]() { // [1] takes by reference
std::cout << "v: " << var << std::endl; // lock mutex for printing
io_service.stop();
});
var = 10; // [2] , lock mutex for writing
// synchronization must be added because between post and t.join calls,
// reading and writing operations are executed
t.join();
in this case you would have to protect var.
answered Nov 12 at 21:49
rafix07
6,6131613
So, are you saying: as thelambdapasses into threadtand get queued and executed successfully, the variable also passes?
– hudac
Nov 13 at 7:41
Ifvaris passed by value to lambda, closure object holds it as data member (copy of thisvar), so while passing closure to thread , we can say that the copy ofvaris passed.
– rafix07
Nov 13 at 7:56
add a comment |
It is thread-safe.
Closure object is created by main thread (with copying var value) after var was created and initialized.
Next, closure object is passed as argument to post method which queues this function object and returns immediately without calling functor. Functor is called between post and t.join calls - post guarantees it.
So your code must be thread-safe.
You would need some synchronization method (for example, use of mutex+lock_guard)
if var was passed by reference [1] and some writing operations on var [2]
were performed between post and t.join calls:
auto var = 1;
io_service.post([&io_service, &var]() { // [1] takes by reference
std::cout << "v: " << var << std::endl; // lock mutex for printing
io_service.stop();
});
var = 10; // [2] , lock mutex for writing
// synchronization must be added because between post and t.join calls,
// reading and writing operations are executed
t.join();
in this case you would have to protect var.
answered Nov 12 at 21:49
rafix07
6,6131613
It is thread-safe.
Closure object is created by main thread (with copying var value) after var was created and initialized.
Next, closure object is passed as argument to post method which queues this function object and returns immediately without calling functor. Functor is called between post and t.join calls - post guarantees it.
So your code must be thread-safe.
You would need some synchronization method (for example, use of mutex+lock_guard)
if var was passed by reference [1] and some writing operations on var [2]
were performed between post and t.join calls:
auto var = 1;
io_service.post([&io_service, &var]() { // [1] takes by reference
std::cout << "v: " << var << std::endl; // lock mutex for printing
io_service.stop();
});
var = 10; // [2] , lock mutex for writing
// synchronization must be added because between post and t.join calls,
// reading and writing operations are executed
t.join();
in this case you would have to protect var.
answered Nov 12 at 21:49
rafix07
6,6131613
answered Nov 12 at 21:49
rafix07
6,6131613
answered Nov 12 at 21:49
rafix07
6,6131613
answered Nov 12 at 21:49
rafix07
6,6131613
6,6131613
So, are you saying: as thelambdapasses into threadtand get queued and executed successfully, the variable also passes?
– hudac
Nov 13 at 7:41
Ifvaris passed by value to lambda, closure object holds it as data member (copy of thisvar), so while passing closure to thread , we can say that the copy ofvaris passed.
– rafix07
Nov 13 at 7:56
add a comment |
So, are you saying: as thelambdapasses into threadtand get queued and executed successfully, the variable also passes?
– hudac
Nov 13 at 7:41
Ifvaris passed by value to lambda, closure object holds it as data member (copy of thisvar), so while passing closure to thread , we can say that the copy ofvaris passed.
– rafix07
Nov 13 at 7:56
So, are you saying: as the
lambda passes into thread t and get queued and executed successfully, the variable also passes?– hudac
Nov 13 at 7:41
So, are you saying: as the
lambda passes into thread t and get queued and executed successfully, the variable also passes?– hudac
Nov 13 at 7:41
If
var is passed by value to lambda, closure object holds it as data member (copy of this var), so while passing closure to thread , we can say that the copy of var is passed.– rafix07
Nov 13 at 7:56
If
var is passed by value to lambda, closure object holds it as data member (copy of this var), so while passing closure to thread , we can say that the copy of var is passed.– rafix07
Nov 13 at 7:56
add a comment |
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