CODEWITHSUNDEEP
c++multithreadingc++11pass-by-referencestdthread
austinmarton
austinmarton

Reputation: 2388

Passing object by reference to std::thread in C++11

Why can't you pass an object by reference when creating a std::thread ?

For example the following snippit gives a compile error:

#include <iostream>
#include <thread>

using namespace std;

static void SimpleThread(int& a)  // compile error
//static void SimpleThread(int a)     // OK
{
    cout << __PRETTY_FUNCTION__ << ":" << a << endl;
}

int main()
{
    int a = 6;

    auto thread1 = std::thread(SimpleThread, a);

    thread1.join();
    return 0;
}

Error:

In file included from /usr/include/c++/4.8/thread:39:0,
                 from ./std_thread_refs.cpp:5:
/usr/include/c++/4.8/functional: In instantiation of ‘struct std::_Bind_simple<void (*(int))(int&)>’:
/usr/include/c++/4.8/thread:137:47:   required from ‘std::thread::thread(_Callable&&, _Args&& ...) [with _Callable = void (&)(int&); _Args = {int&}]’
./std_thread_refs.cpp:19:47:   required from here
/usr/include/c++/4.8/functional:1697:61: error: no type named ‘type’ in ‘class std::result_of<void (*(int))(int&)>’
       typedef typename result_of<_Callable(_Args...)>::type result_type;
                                                             ^
/usr/include/c++/4.8/functional:1727:9: error: no type named ‘type’ in ‘class std::result_of<void (*(int))(int&)>’
         _M_invoke(_Index_tuple<_Indices...>)
         ^

I've changed to passing a pointer, but is there a better work around?

Upvotes: 114

Views: 77239

Answers (3)

Jarod42
Jarod42

Reputation: 218288

std::thread copy(/move) its arguments, you might even see the note:

The arguments to the thread function are moved or copied by value. If a reference argument needs to be passed to the thread function, it has to be wrapped (e.g., with std::ref or std::cref).

So, you might use std::reference_wrapper through std::ref/std::cref:

auto thread1 = std::thread(SimpleThread, std::ref(a));

or use lambda:

auto thread1 = std::thread([&a]() { SimpleThread(a); });

Upvotes: 11

jfMR
jfMR

Reputation: 24788

Based on this comment, this answer elaborates on the reason why the arguments are not passed by reference to the thread function by default.

Consider the following function SimpleThread():

void SimpleThread(int& i) {
    std::this_thread::sleep_for(std::chrono::seconds{1});
    i = 0;
}

Now, imagine what would happen if the following code compiled (it does not compile):

int main()
{
    {
        int a;
        std::thread th(SimpleThread, a);
        th.detach();
    }
    // "a" is out of scope
    // at this point the thread may be still running
    // ...
}

The argument a would be passed by reference to SimpleThread(). The thread may still be sleeping in the function SimpleThread() after the variable a has already gone out of scope and its lifetime has ended. If so, i in SimpleThread() would actually be a dangling reference, and the assignment i = 0 would result in undefined behaviour.

By wrapping reference arguments with the class template std::reference_wrapper (using the function templates std::ref and std::cref) you explicitly express your intentions.

Upvotes: 40

ShadowRanger
ShadowRanger

Reputation: 155684

Explicitly initialize the thread with a reference_wrapper by using std::ref:

auto thread1 = std::thread(SimpleThread, std::ref(a));

(or std::cref instead of std::ref, as appropriate). Per notes from cppreference on std:thread:

The arguments to the thread function are moved or copied by value. If a reference argument needs to be passed to the thread function, it has to be wrapped (e.g. with std::ref or std::cref).

Upvotes: 150

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