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Is there a way to make member function NOT callable from constructor?


What are the differences between a pointer variable and a reference variable in C++?Can I call a constructor from another constructor (do constructor chaining) in C++?Throwing exceptions from constructorsHow do I call ::std::make_shared on a class with only protected or private constructors?Calling a base member in constructor in multiple inheritance in C++Equality-compare std::weak_ptrClass inheritance: Constructor and member functions of class not recognized by compilerHow does shared_ptr<T> detect that T derives from enable_shared_from_this<T>?enable_shared_from_this derived class methods are undefined referenceDefault move constructor with mutex member






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9















I have member function (method) which uses



std::enable_shared_from_this::weak_from_this() 


In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.



Is there a way to check against it at compile time?










share|improve this question
























  • Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.

    – rubenvb
    5 hours ago






  • 1





    Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.

    – SergeyA
    5 hours ago

















9















I have member function (method) which uses



std::enable_shared_from_this::weak_from_this() 


In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.



Is there a way to check against it at compile time?










share|improve this question
























  • Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.

    – rubenvb
    5 hours ago






  • 1





    Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.

    – SergeyA
    5 hours ago













9












9








9


1






I have member function (method) which uses



std::enable_shared_from_this::weak_from_this() 


In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.



Is there a way to check against it at compile time?










share|improve this question
















I have member function (method) which uses



std::enable_shared_from_this::weak_from_this() 


In short: weak_from_this returns weak_ptr to this. One caveat is it can't be used from constructor.
If somebody would use my function from constructor of inherited class, weak_from_this inside it would return expired weak_ptr. I guard against that with assertion checking that it's not expired, but it's a run-time check.



Is there a way to check against it at compile time?







c++ c++17 shared-ptr weak-ptr






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 5 hours ago









armitus

509114




509114










asked 5 hours ago









KorriKorri

32627




32627












  • Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.

    – rubenvb
    5 hours ago






  • 1





    Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.

    – SergeyA
    5 hours ago

















  • Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.

    – rubenvb
    5 hours ago






  • 1





    Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.

    – SergeyA
    5 hours ago
















Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.

– rubenvb
5 hours ago





Note there is a difference in scope between a child class constructor body and the parent class constructor: the latter has been executed completely before you even start initializing the child class's members (if any), let alone enter the child class constructor body.

– rubenvb
5 hours ago




1




1





Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.

– SergeyA
5 hours ago





Nice question. One way would be to make a dummy class with pure virtual function weak_from_this and inherit yours from it. This will make it a hard compile error.

– SergeyA
5 hours ago












3 Answers
3






active

oldest

votes


















5














I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.






share|improve this answer






























    2














    No there is no way. Consider:



    void call_me(struct widget*);

    struct widget : std::enable_shared_from_this<widget>
    widget()
    call_me(this);


    void display()
    shared_from_this();

    ;

    // later:

    void call_me(widget* w)
    w->display(); // crash



    The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.






    share|improve this answer






























      2














      Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:



      class A 

      // ... whatever ...

      A()
      // do construction work
      constructed = true;


      foo()
      if (not constructed)
      throw std::logic_error("Cannot call foo() during construction");

      // the rest of foo


      bool constructed false ;



      You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.



      An alternative to throwing could be assert()'ing.






      share|improve this answer























      • Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

        – Korri
        2 hours ago











      • @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

        – Jesper Juhl
        1 hour ago












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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5














      I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.






      share|improve this answer



























        5














        I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.






        share|improve this answer

























          5












          5








          5







          I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.






          share|improve this answer













          I am afraid the answer is "no, it's not possible to protect against this at compile-time." It's always difficult to prove a negative, but consider this: if it was possible to protect a function this way, it would probably have been done for weak_from_this and shared_from_this in the standard library itself.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 5 hours ago









          AngewAngew

          134k11260353




          134k11260353























              2














              No there is no way. Consider:



              void call_me(struct widget*);

              struct widget : std::enable_shared_from_this<widget>
              widget()
              call_me(this);


              void display()
              shared_from_this();

              ;

              // later:

              void call_me(widget* w)
              w->display(); // crash



              The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.






              share|improve this answer



























                2














                No there is no way. Consider:



                void call_me(struct widget*);

                struct widget : std::enable_shared_from_this<widget>
                widget()
                call_me(this);


                void display()
                shared_from_this();

                ;

                // later:

                void call_me(widget* w)
                w->display(); // crash



                The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.






                share|improve this answer

























                  2












                  2








                  2







                  No there is no way. Consider:



                  void call_me(struct widget*);

                  struct widget : std::enable_shared_from_this<widget>
                  widget()
                  call_me(this);


                  void display()
                  shared_from_this();

                  ;

                  // later:

                  void call_me(widget* w)
                  w->display(); // crash



                  The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.






                  share|improve this answer













                  No there is no way. Consider:



                  void call_me(struct widget*);

                  struct widget : std::enable_shared_from_this<widget>
                  widget()
                  call_me(this);


                  void display()
                  shared_from_this();

                  ;

                  // later:

                  void call_me(widget* w)
                  w->display(); // crash



                  The thing is there is a reason you want to check for not calling shared_from_this in the constructor. Think about that reason. It's not that shared_from_this cannot be called, it's because it's return value has no way of being assigned yet. It is also not because it will never be assigned. It's because it will be assigned later in the execution of the code. Order of operation is a runtime property of your program. You cannot assert at compile time for order of operation, which is done at runtime.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 4 hours ago









                  Guillaume RacicotGuillaume Racicot

                  16.3k53872




                  16.3k53872





















                      2














                      Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:



                      class A 

                      // ... whatever ...

                      A()
                      // do construction work
                      constructed = true;


                      foo()
                      if (not constructed)
                      throw std::logic_error("Cannot call foo() during construction");

                      // the rest of foo


                      bool constructed false ;



                      You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.



                      An alternative to throwing could be assert()'ing.






                      share|improve this answer























                      • Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

                        – Korri
                        2 hours ago











                      • @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

                        – Jesper Juhl
                        1 hour ago
















                      2














                      Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:



                      class A 

                      // ... whatever ...

                      A()
                      // do construction work
                      constructed = true;


                      foo()
                      if (not constructed)
                      throw std::logic_error("Cannot call foo() during construction");

                      // the rest of foo


                      bool constructed false ;



                      You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.



                      An alternative to throwing could be assert()'ing.






                      share|improve this answer























                      • Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

                        – Korri
                        2 hours ago











                      • @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

                        – Jesper Juhl
                        1 hour ago














                      2












                      2








                      2







                      Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:



                      class A 

                      // ... whatever ...

                      A()
                      // do construction work
                      constructed = true;


                      foo()
                      if (not constructed)
                      throw std::logic_error("Cannot call foo() during construction");

                      // the rest of foo


                      bool constructed false ;



                      You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.



                      An alternative to throwing could be assert()'ing.






                      share|improve this answer













                      Not as such, but - if performance is not an issue, you could add a flag which indicates construction is complete, and use that to fail at run-time with such calls:



                      class A 

                      // ... whatever ...

                      A()
                      // do construction work
                      constructed = true;


                      foo()
                      if (not constructed)
                      throw std::logic_error("Cannot call foo() during construction");

                      // the rest of foo


                      bool constructed false ;



                      You could also make these checks only apply when compiling in DEBUG mode (e.g. with conditional compilation using the preprocessor - #ifndef NDEBUG) so that at run time you won't get the performance penalty. Mind the noexcepts though.



                      An alternative to throwing could be assert()'ing.







                      share|improve this answer












                      share|improve this answer



                      share|improve this answer










                      answered 3 hours ago









                      einpoklumeinpoklum

                      37k28132263




                      37k28132263












                      • Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

                        – Korri
                        2 hours ago











                      • @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

                        – Jesper Juhl
                        1 hour ago


















                      • Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

                        – Korri
                        2 hours ago











                      • @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

                        – Jesper Juhl
                        1 hour ago

















                      Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

                      – Korri
                      2 hours ago





                      Since apparently there isn't compile-time solution which doesn't make code less readable, I decided to go with assert(!wptr.expired()). I think it's a little bit more fitting than exception, because there is no way to recover from this situation.

                      – Korri
                      2 hours ago













                      @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

                      – Jesper Juhl
                      1 hour ago






                      @Korri remember that asserts are usually compiled out in release builds, so nothing will happen. An exception however is not, so it would still terminate the program (if not caught and swallowed) in a release build. You could do both; first assert then throw, or just throw.

                      – Jesper Juhl
                      1 hour ago


















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                      Are there any AGPL-style licences that require source code modifications to be public? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

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                      Mortes em março de 2019 Referências Menu de navegação«Zhores Alferov, Nobel de Física bielorrusso, morre aos 88 anos - Ciência»«Fallece Rafael Torija, o bispo emérito de Ciudad Real»«Peter Hurford dies at 88»«Keith Flint, vocalista do The Prodigy, morre aos 49 anos»«Luke Perry, ator de 'Barrados no baile' e 'Riverdale', morre aos 52 anos»«Former Rangers and Scotland captain Eric Caldow dies, aged 84»«Morreu, aos 61 anos, a antiga lenda do wrestling King Kong Bundy»«Fallece el actor y director teatral Abraham Stavans»«In Memoriam Guillaume Faye»«Sidney Sheinberg, a Force Behind Universal and Spielberg, Is Dead at 84»«Carmine Persico, Colombo Crime Family Boss, Is Dead at 85»«Dirigent Michael Gielen gestorben»«Ciclista tricampeã mundial e prata na Rio 2016 é encontrada morta em casa aos 23 anos»«Pagan Community Notes: Raven Grimassi dies, Indianapolis pop-up event cancelled, Circle Sanctuary announces new podcast, and more!»«Hal Blaine, Wrecking Crew Drummer, Dies at 90»«Morre Coutinho, que editou dupla lendária com Pelé no Santos»«Cantor Demétrius, ídolo da Jovem Guarda, morre em SP»«Ex-presidente do Vasco, Eurico Miranda morre no Rio de Janeiro»«Bronze no Mundial de basquete de 1971, Laís Elena morre aos 76 anos»«Diretor de Corridas da F1, Charlie Whiting morre aos 66 anos às vésperas do GP da Austrália»«Morreu o cardeal Danneels, da Bélgica»«Morreu o cartoonista Augusto Cid»«Morreu a atriz Maria Isabel de Lizandra, de "Vale Tudo" e novelas da Tupi»«WS Merwin, prize-winning poet of nature, dies at 91»«Atriz Márcia Real morre em São Paulo aos 88 anos»«Mauritanie: décès de l'ancien président Mohamed Mahmoud ould Louly»«Morreu Dick Dale, o rei da surf guitar e de "Pulp Fiction"»«Falleció Víctor Genes»«João Carlos Marinho, autor de 'O Gênio do Crime', morre em SP»«Legendary Horror Director and SFX Artist John Carl Buechler Dies at 66»«Morre em Salvador a religiosa Makota Valdina»«مرگ بازیکن‌ سابق نساجی بر اثر سقوط سنگ در مازندران»«Domingos Oliveira morre no Rio»«Morre Airton Ravagniani, ex-São Paulo, Fla, Vasco, Grêmio e Sport - Notícias»«Morre o escritor Flavio Moreira da Costa»«Larry Cohen, Writer-Director of 'It's Alive' and 'Hell Up in Harlem,' Dies at 77»«Scott Walker, experimental singer-songwriter, dead at 76»«Joseph Pilato, Day of the Dead Star and Horror Favorite, Dies at 70»«Sheffield United set to pay tribute to legendary goalkeeper Ted Burgin who has died at 91»«Morre Rafael Henzel, sobrevivente de acidente aéreo da Chapecoense»«Morre Valery Bykovsky, um dos primeiros cosmonautas da União Soviética»«Agnès Varda, cineasta da Nouvelle Vague, morre aos 90 anos»«Agnès Varda, cineasta francesa, morre aos 90 anos»«Tania Mallet, James Bond Actress and Helen Mirren's Cousin, Dies at 77»e