Start thread with member function

Question

I am trying to construct a std  thread with a member function that takes no arguments and returns void  I can t figure out any syntax that works - the compiler complains no matter what  What is the correct way to implement spawn   so that it returns a std  thread that executes test      include  lt thread gt  class blub     void test         public    std  thread spawn         return   test

User · Answer

Since you are using C  11  lambda-expression is a nice amp clean solution  class blub       void test        public      std  thread spawn           return std  thread   this    this- gt test                   since this- gt  can be omitted  it could be shorten to  std  thread   this    test         or just  deprecated  std  thread        test

User · Answer

Here is a complete example   include  lt thread gt   include  lt iostream gt   class Wrapper      public        void member1               std  cout  lt  lt   i am member1   lt  lt  std  endl                void member2 const char  arg1  unsigned arg2              std  cout  lt  lt   i am member2 and my first arg is     lt  lt  arg1  lt  lt     and second arg is     lt  lt  arg2  lt  lt       lt  lt  std  endl                std  thread member1Thread               return std  thread       member1                      std  thread member2Thread const char  arg1  unsigned arg2              return std  thread       member2 arg1  arg2                  int main int argc  char   argv       Wrapper  w   new Wrapper       std  thread tw1   w- gt member1Thread       std  thread tw2   w- gt member2Thread  hello   100      tw1 join       tw2 join       return 0      Compiling with g   produces the following result  g   -Wall -std c  11 hello cc -o hello -pthread  i am member1 i am member2 and my first arg is  hello  and second arg is  100

User · Answer

Some users have already given their answer and explained it very well   I would like to add few more things related to thread    How to work with functor and thread   Please refer to below example  The thread will make its own copy of the object while passing the object    include lt thread gt   include lt Windows h gt   include lt iostream gt   using namespace std   class CB    public      CB                 cout  lt  lt   this    lt  lt  this  lt  lt  endl            void operator          void CB  operator           cout  lt  lt   this    lt  lt  this  lt  lt  endl      for  int i   0  i  lt  5  i                  cout  lt  lt   CB      lt  lt  i  lt  lt  endl          Sleep 1000            void main         CB obj         please note the address of obj       thread t obj      here obj will be passed by value                       i e  thread will make it own local copy of it                         we can confirm it by matching the address of                       object printed in the constructor                        and address of the obj printed in the function      t join         Another way of achieving the same thing is like   void main         thread t  CB           t join        But if you want to pass the object by reference then use the below syntax   void main         CB obj        thread t obj       thread t std  ref obj        t join

User · Answer

hop5 and  RnMss suggested to use C  11 lambdas  but if you deal with pointers  you can use them directly    include  lt thread gt   include  lt iostream gt   class CFoo     public      int m i   0      void bar             m i            int main       CFoo foo    std  thread t1  amp CFoo  bar   amp foo     t1 join      std  thread t2  amp CFoo  bar   amp foo     t2 join      std  cout  lt  lt  foo m i  lt  lt  std  endl    return 0      outputs  2   Rewritten sample from this answer would be then    include  lt thread gt   include  lt iostream gt   class Wrapper     public        void member1               std  cout  lt  lt   i am member1   lt  lt  std  endl                void member2 const char  arg1  unsigned arg2              std  cout  lt  lt   i am member2 and my first arg is     lt  lt  arg1  lt  lt     and second arg is     lt  lt  arg2  lt  lt       lt  lt  std  endl                std  thread member1Thread               return std  thread  amp Wrapper  member1  this                 std  thread member2Thread const char  arg1  unsigned arg2              return std  thread  amp Wrapper  member2  this  arg1  arg2               int main       Wrapper  w   new Wrapper      std  thread tw1   w- gt member1Thread      tw1 join      std  thread tw2   w- gt member2Thread  hello   100     tw2 join      return 0

User · Answer

include  lt thread gt   include  lt iostream gt   class bar   public    void foo         std  cout  lt  lt   hello from member function   lt  lt  std  endl          int main       std  thread t  amp bar  foo  bar       t join        EDIT  Accounting your edit  you have to do it like this     std  thread spawn         return std  thread  amp blub  test  this           UPDATE  I want to explain some more points  some of them have also been discussed in the comments   The syntax described above is defined in terms of the INVOKE definition    20 8 2 1       Define INVOKE  f  t1  t2       tN  as follows            t1  f  t2       tN  when f is a pointer to a member function of a class T and t1 is an object of type T or a reference to an object of   type T or a reference to an object of a type derived from T       t1   f  t2       tN  when f is a pointer to a member function of a class T and t1 is not one of the types described in the previous   item    t1  f when N       1 and f is a pointer to member data of a class T and  t  1 is an object of type T or a   reference to an object of type T or a reference to an object of a   type derived from T      t1   f when N    1 and f is a pointer to member data of a class T and t 1 is not one of the types described in the previous item    f t1  t2       tN  in all other cases         Another general fact which I want to point out is that by default the thread constructor will copy all arguments passed to it  The reason for this is that the arguments may need to outlive the calling thread  copying the arguments guarantees that  Instead  if you want to really pass a reference  you can use a std  reference wrapper created by std  ref   std  thread  foo  std  ref arg1      By doing this  you are promising that you will take care of guaranteeing that the arguments will still exist when the thread operates on them     Note that all the things mentioned above can also be applied to std  async and std  bind

[c++] Start thread with member function

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