boost:shared_ptr cast problem

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  • Jun
    #1

    boost:shared_ptr cast problem

    Hello,

    I've code like :
    =============== =============== =============
    class A{
    public :
    // create print content
    friend std::ostream& operator<< (std::ostream&
    os, const A& a);

    }

    typedef boost::shared_p tr<AAPtr; // Define A class smart pointer

    APtr aPtr1(new A());
    APtr aPtr2(new A());
    APtr aPtr3(new A());

    vector<APtrAVec ;
    AVec.push_back( aPtr1);
    AVec.push_back( aPtr2);
    AVec.push_back( aPtr3);

    std::copy(AVec. begin(),AVec.en d(),
    std::ostream_it erator<APtr>(st d::cout, "\n"));

    std::copy(AVec. begin(),AVec.en d(), std::ostream_it erator< share_ptr<A>
    >(std::cout, "\n"));
    =============== =============== =============

    The two copy methods only output the address of pointers, they could
    just print the class A defined print contents. And following code
    works :
    =============== =============== =============
    vector<APtr>::i terator i = AVec.begin();
    for(; i!= AVec.end(); ++i){
    cout << *static_cast<AP tr>(*i) << endl;
    }
    =============== =============== =============

    Anyone has some ideas for that ? Thank you in advance.


    Jun
    Tags: None
    • Jun
      #2
      Re: boost:shared_pt r cast problem

      On Nov 25, 10:12 pm, Jun <junh...@gmail. comwrote:
      Hello,
      >
      I've code like :
      =============== =============== =============
      class A{
      public :
      // create print content
      friend std::ostream& operator<< (std::ostream&
      os, const A& a);
      >
      }
      >
      typedef boost::shared_p tr<AAPtr; // Define A class smart pointer
      >
      APtr aPtr1(new A());
      APtr aPtr2(new A());
      APtr aPtr3(new A());
      >
      vector<APtrAVec ;
      AVec.push_back( aPtr1);
      AVec.push_back( aPtr2);
      AVec.push_back( aPtr3);
      >
      std::copy(AVec. begin(),AVec.en d(),
      std::ostream_it erator<APtr>(st d::cout, "\n"));
      >
      std::copy(AVec. begin(),AVec.en d(), std::ostream_it erator< share_ptr<A>>(s td::cout, "\n"));
      >
      =============== =============== =============
      >
      The two copy methods only output the address of pointers, they could
      just print the class A defined print contents. And following code
      works :
      =============== =============== =============
      vector<APtr>::i terator i = AVec.begin();
      for(; i!= AVec.end(); ++i){
      cout << *static_cast<AP tr>(*i) << endl;
      }
      =============== =============== =============
      >
      Anyone has some ideas for that ? Thank you in advance.
      >
      Jun
      Actually, It's boost::shared_p tr serialization problem.

        Comment

        • Kai-Uwe Bux
          #3
          Re: boost:shared_pt r cast problem

          Jun wrote:
          Hello,
          >
          I've code like :
          =============== =============== =============
          class A{
          public :
          // create print content
          friend std::ostream& operator<< (std::ostream&
          os, const A& a);
          >
          }
          >
          typedef boost::shared_p tr<AAPtr; // Define A class smart pointer
          >
          APtr aPtr1(new A());
          APtr aPtr2(new A());
          APtr aPtr3(new A());
          >
          vector<APtrAVec ;
          AVec.push_back( aPtr1);
          AVec.push_back( aPtr2);
          AVec.push_back( aPtr3);
          >
          std::copy(AVec. begin(),AVec.en d(),
          std::ostream_it erator<APtr>(st d::cout, "\n"));
          >
          std::copy(AVec. begin(),AVec.en d(),
          std::ostream_it erator< share_ptr<A(std ::cout, "\n"));
          =============== =============== =============
          >
          The two copy methods only output the address of pointers, they could
          just print the class A defined print contents.
          Why would you expect that? Consider

          int i = 5;
          std::cout << i << std::endl
          << &i << std::endl;

          You would expect the second line to print an address, would you not? Why
          should shared_ptr<beha ve differently? Also note that shared_ptr could
          have null value. In that case, there is no pointee that could be printed
          instead of an address.

          And following code works :
          =============== =============== =============
          vector<APtr>::i terator i = AVec.begin();
          for(; i!= AVec.end(); ++i){
          cout << *static_cast<AP tr>(*i) << endl;
          }
          =============== =============== =============
          Anyone has some ideas for that ?
          Yes, leave out the cast: *i is already of type APtr.

          cout << *(*i) << endl;



          BTW: I have the feeling that I did not really understand what the real
          problem is that you want to solve. Maybe, you oversimplified it for the
          purpose of the post. Could you provide a little background as to _why_ you
          want to print the shared_ptr and _why_ you feel it would be the
          RightThing(tm) if that printed the pointee instead of an address? Maybe, it
          is the context of the underlying problem that makes you think printing the
          pointee would be right. In that case, it would be good to share the
          underlying problem with us, because it might have a well known solution.


          Best

          Kai-Uwe Bux

            Comment

            • Jun
              #4
              Re: boost:shared_pt r cast problem

              On Nov 26, 1:52 am, Kai-Uwe Bux <jkherci...@gmx .netwrote:
              Jun wrote:
              Hello,
              >
              I've code like :
              =============== =============== =============
              class A{
              public :
              // create print content
              friend std::ostream& operator<< (std::ostream&
              os, const A& a);
              >
              }
              >
              typedef boost::shared_p tr<AAPtr; // Define A class smart pointer
              >
              APtr aPtr1(new A());
              APtr aPtr2(new A());
              APtr aPtr3(new A());
              >
              vector<APtrAVec ;
              AVec.push_back( aPtr1);
              AVec.push_back( aPtr2);
              AVec.push_back( aPtr3);
              >
              std::copy(AVec. begin(),AVec.en d(),
              std::ostream_it erator<APtr>(st d::cout, "\n"));
              >
              std::copy(AVec. begin(),AVec.en d(),
              std::ostream_it erator< share_ptr<A(std ::cout, "\n"));
              =============== =============== =============
              >
              The two copy methods only output the address of pointers, they could
              just print the class A defined print contents.
              >
              Why would you expect that? Consider
              >
              int i = 5;
              std::cout << i << std::endl
              << &i << std::endl;
              >
              You would expect the second line to print an address, would you not? Why
              should shared_ptr<beha ve differently? Also note that shared_ptr could
              have null value. In that case, there is no pointee that could be printed
              instead of an address.
              >
              And following code works :
              =============== =============== =============
              vector<APtr>::i terator i = AVec.begin();
              for(; i!= AVec.end(); ++i){
              cout << *static_cast<AP tr>(*i) << endl;
              }
              =============== =============== =============
              Anyone has some ideas for that ?
              >
              Yes, leave out the cast: *i is already of type APtr.
              >
              cout << *(*i) << endl;
              >
              BTW: I have the feeling that I did not really understand what the real
              problem is that you want to solve. Maybe, you oversimplified it for the
              purpose of the post. Could you provide a little background as to _why_ you
              want to print the shared_ptr and _why_ you feel it would be the
              RightThing(tm) if that printed the pointee instead of an address? Maybe, it
              is the context of the underlying problem that makes you think printing the
              pointee would be right. In that case, it would be good to share the
              underlying problem with us, because it might have a well known solution.
              >
              Best
              >
              Kai-Uwe Bux
              I've a person class, which contains name, age. I applied output as
              name age,
              for my custom class. By storing it as a smart pointer PersonPtr in a
              vector.
              Then my idea is using copy algorithm to print out all the PersonPtr in
              the
              vector, but failed. Since PersonPtr has the serialization problem. By
              using
              copy algorithm, it only prints the address of smart pointer instead
              name age,
              which i defined.

                Comment

                • Kai-Uwe Bux
                  #5
                  Re: boost:shared_pt r cast problem

                  Jun wrote:
                  On Nov 26, 1:52 am, Kai-Uwe Bux <jkherci...@gmx .netwrote:
                  >Jun wrote:
                  Hello,
                  >>
                  I've code like :
                  =============== =============== =============
                  class A{
                  public :
                  // create print content
                  friend std::ostream& operator<< (std::ostream&
                  os, const A& a);
                  >>
                  }
                  >>
                  typedef boost::shared_p tr<AAPtr; // Define A class smart pointer
                  >>
                  APtr aPtr1(new A());
                  APtr aPtr2(new A());
                  APtr aPtr3(new A());
                  >>
                  vector<APtrAVec ;
                  AVec.push_back( aPtr1);
                  AVec.push_back( aPtr2);
                  AVec.push_back( aPtr3);
                  >>
                  std::copy(AVec. begin(),AVec.en d(),
                  std::ostream_it erator<APtr>(st d::cout, "\n"));
                  >>
                  std::copy(AVec. begin(),AVec.en d(),
                  std::ostream_it erator< share_ptr<A(std ::cout, "\n"));
                  =============== =============== =============
                  >>
                  The two copy methods only output the address of pointers, they could
                  just print the class A defined print contents.
                  >>
                  >Why would you expect that? Consider
                  >>
                  > int i = 5;
                  > std::cout << i << std::endl
                  > << &i << std::endl;
                  >>
                  >You would expect the second line to print an address, would you not? Why
                  >should shared_ptr<beha ve differently? Also note that shared_ptr could
                  >have null value. In that case, there is no pointee that could be printed
                  >instead of an address.
                  >>
                  And following code works :
                  =============== =============== =============
                  vector<APtr>::i terator i = AVec.begin();
                  for(; i!= AVec.end(); ++i){
                  cout << *static_cast<AP tr>(*i) << endl;
                  }
                  =============== =============== =============
                  Anyone has some ideas for that ?
                  >>
                  >Yes, leave out the cast: *i is already of type APtr.
                  >>
                  > cout << *(*i) << endl;
                  >>
                  >BTW: I have the feeling that I did not really understand what the real
                  >problem is that you want to solve. Maybe, you oversimplified it for the
                  >purpose of the post. Could you provide a little background as to _why_
                  >you want to print the shared_ptr and _why_ you feel it would be the
                  >RightThing(t m) if that printed the pointee instead of an address? Maybe,
                  >it is the context of the underlying problem that makes you think printing
                  >the pointee would be right. In that case, it would be good to share the
                  >underlying problem with us, because it might have a well known solution.
                  >>
                  >Best
                  >>
                  >Kai-Uwe Bux
                  >
                  I've a person class, which contains name, age. I applied output as
                  name age,
                  for my custom class. By storing it as a smart pointer PersonPtr in a
                  vector.
                  Then my idea is using copy algorithm to print out all the PersonPtr in
                  the
                  vector, but failed. Since PersonPtr has the serialization problem. By
                  using
                  copy algorithm, it only prints the address of smart pointer instead
                  name age,
                  which i defined.
                  You might try:

                  template < typename T >
                  T const & deref ( std::tr1::share d_ptr<Tptr ) {
                  return ( *ptr );
                  }

                  and

                  std::transform( AVec.begin(), AVec.end(),
                  std::ostream_it erator< A >( std::cout, "\n" ),
                  &deref<A);

                  Maybe, there is a way to use boost::lambda and just say

                  std::transform( AVec.begin(), AVec.end(),
                  std::ostream_it erator< A >( std::cout, "\n" ),
                  *_1);

                  without the need to define deref.


                  Best

                  Kai-Uwe Bux

                    Comment

                    • Jun
                      #6
                      Re: boost:shared_pt r cast problem

                      On Nov 26, 1:10 pm, Kai-Uwe Bux <jkherci...@gmx .netwrote:
                      Jun wrote:
                      On Nov 26, 1:52 am, Kai-Uwe Bux <jkherci...@gmx .netwrote:
                      Jun wrote:
                      Hello,
                      >
                      I've code like :
                      =============== =============== =============
                      class A{
                      public :
                      // create print content
                      friend std::ostream& operator<< (std::ostream&
                      os, const A& a);
                      >
                      }
                      >
                      typedef boost::shared_p tr<AAPtr; // Define A class smart pointer
                      >
                      APtr aPtr1(new A());
                      APtr aPtr2(new A());
                      APtr aPtr3(new A());
                      >
                      vector<APtrAVec ;
                      AVec.push_back( aPtr1);
                      AVec.push_back( aPtr2);
                      AVec.push_back( aPtr3);
                      >
                      std::copy(AVec. begin(),AVec.en d(),
                      std::ostream_it erator<APtr>(st d::cout, "\n"));
                      >
                      std::copy(AVec. begin(),AVec.en d(),
                      std::ostream_it erator< share_ptr<A(std ::cout, "\n"));
                      =============== =============== =============
                      >
                      The two copy methods only output the address of pointers, they could
                      just print the class A defined print contents.
                      >
                      Why would you expect that? Consider
                      >
                      int i = 5;
                      std::cout << i << std::endl
                      << &i << std::endl;
                      >
                      You would expect the second line to print an address, would you not? Why
                      should shared_ptr<beha ve differently? Also note that shared_ptr could
                      have null value. In that case, there is no pointee that could be printed
                      instead of an address.
                      >
                      And following code works :
                      =============== =============== =============
                      vector<APtr>::i terator i = AVec.begin();
                      for(; i!= AVec.end(); ++i){
                      cout << *static_cast<AP tr>(*i) << endl;
                      }
                      =============== =============== =============
                      Anyone has some ideas for that ?
                      >
                      Yes, leave out the cast: *i is already of type APtr.
                      >
                      cout << *(*i) << endl;
                      >
                      BTW: I have the feeling that I did not really understand what the real
                      problem is that you want to solve. Maybe, you oversimplified it for the
                      purpose of the post. Could you provide a little background as to _why_
                      you want to print the shared_ptr and _why_ you feel it would be the
                      RightThing(tm) if that printed the pointee instead of an address? Maybe,
                      it is the context of the underlying problem that makes you think printing
                      the pointee would be right. In that case, it would be good to share the
                      underlying problem with us, because it might have a well known solution.
                      >
                      Best
                      >
                      Kai-Uwe Bux
                      >
                      I've a person class, which contains name, age. I applied output as
                      name age,
                      for my custom class. By storing it as a smart pointer PersonPtr in a
                      vector.
                      Then my idea is using copy algorithm to print out all the PersonPtr in
                      the
                      vector, but failed. Since PersonPtr has the serialization problem. By
                      using
                      copy algorithm, it only prints the address of smart pointer instead
                      name age,
                      which i defined.
                      >
                      You might try:
                      >
                      template < typename T >
                      T const & deref ( std::tr1::share d_ptr<Tptr ) {
                      return ( *ptr );
                      >
                      }
                      >
                      and
                      >
                      std::transform( AVec.begin(), AVec.end(),
                      std::ostream_it erator< A >( std::cout, "\n" ),
                      &deref<A);
                      >
                      Maybe, there is a way to use boost::lambda and just say
                      >
                      std::transform( AVec.begin(), AVec.end(),
                      std::ostream_it erator< A >( std::cout, "\n" ),
                      *_1);
                      >
                      without the need to define deref.
                      >
                      Best
                      >
                      Kai-Uwe Bux
                      I saw the lambda solution, anyway, I will try and post the results.
                      Jun

                        Comment

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