User-defined conversions sequence

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Before I studied the explicit keyword, my teacher said: "compiler doesn't execute consecutive user defined conversion". If it is true, are there any errors in my code? Or have I misunderstood my teacher? I'm working in VS2017.



#include<iostream>
#include <string>

class Myclass
public:
Myclass()
std::cout << "Myclass" << std::endl;

;

class Myclass1
public:
Myclass1(Myclass m)
std::cout << "Myclass1" << std::endl;

;
class Myclass2
public:
Myclass2(Myclass1 m)
std::cout << "Myclass2" << std::endl;

;

int main()
Myclass2 m2 = Myclass;










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    up vote
    14
    down vote

    favorite












    Before I studied the explicit keyword, my teacher said: "compiler doesn't execute consecutive user defined conversion". If it is true, are there any errors in my code? Or have I misunderstood my teacher? I'm working in VS2017.



    #include<iostream>
    #include <string>

    class Myclass
    public:
    Myclass()
    std::cout << "Myclass" << std::endl;

    ;

    class Myclass1
    public:
    Myclass1(Myclass m)
    std::cout << "Myclass1" << std::endl;

    ;
    class Myclass2
    public:
    Myclass2(Myclass1 m)
    std::cout << "Myclass2" << std::endl;

    ;

    int main()
    Myclass2 m2 = Myclass;










    share|improve this question









    New contributor




    User8500049 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      up vote
      14
      down vote

      favorite









      up vote
      14
      down vote

      favorite











      Before I studied the explicit keyword, my teacher said: "compiler doesn't execute consecutive user defined conversion". If it is true, are there any errors in my code? Or have I misunderstood my teacher? I'm working in VS2017.



      #include<iostream>
      #include <string>

      class Myclass
      public:
      Myclass()
      std::cout << "Myclass" << std::endl;

      ;

      class Myclass1
      public:
      Myclass1(Myclass m)
      std::cout << "Myclass1" << std::endl;

      ;
      class Myclass2
      public:
      Myclass2(Myclass1 m)
      std::cout << "Myclass2" << std::endl;

      ;

      int main()
      Myclass2 m2 = Myclass;










      share|improve this question









      New contributor




      User8500049 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      Before I studied the explicit keyword, my teacher said: "compiler doesn't execute consecutive user defined conversion". If it is true, are there any errors in my code? Or have I misunderstood my teacher? I'm working in VS2017.



      #include<iostream>
      #include <string>

      class Myclass
      public:
      Myclass()
      std::cout << "Myclass" << std::endl;

      ;

      class Myclass1
      public:
      Myclass1(Myclass m)
      std::cout << "Myclass1" << std::endl;

      ;
      class Myclass2
      public:
      Myclass2(Myclass1 m)
      std::cout << "Myclass2" << std::endl;

      ;

      int main()
      Myclass2 m2 = Myclass;







      c++ type-conversion implicit-conversion






      share|improve this question









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      Check out our Code of Conduct.











      share|improve this question









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      share|improve this question




      share|improve this question








      edited yesterday









      Evg

      3,69221334




      3,69221334






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      asked yesterday









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      864




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          3 Answers
          3






          active

          oldest

          votes

















          up vote
          10
          down vote



          accepted











          compiler doesn't execute consecutive user defined conversion




          Your teacher is right. In your code sample it means Myclass cannot be converted to Myclass1 when you assign in:



          Myclass2 m2 = Myclass;


          Because constructor expects Myclass1 when creating Myclass2, and compiler cannot consecutively convert Myclass to Myclass1 and then use it for creating Myclass2. But if you have following line:



          Myclass1 m2 = Myclass;


          It will work, because constructor of Myclass1 takes Myclass as argument.



          Update:



          You may ask why this works:



          Myclass2 m2 Myclass;


          Because in this case, constructor is called and conversion can be done implicitly unless you declare Myclass1 as explicit which will fail code compilation (Thanks Fureeish for reminder), but in:



          Myclass2 m2 = Myclass;


          is like calling copy-constructor which needs reference. so if you write it like this, it will work:



          Myclass2 m2 = Myclass1(Myclass);


          As EVG mentioned, Myclass2 m2 = Myclass; is accepted by VS 2017 if the conformance mode (/permissive-) is not activated.






          share|improve this answer


















          • 2




            Worth noting that OP mentioned explicit. In your first example in your edit, there is a temporary Myclass1 object created implicitely from Myclass instance (that's why there is a Myclass2 constructor invoked, which requires Myclass1 object). If you make Myclass1's constructor explicit, it will fail
            – Fureeish
            yesterday











          • @Fureeish I will add your comment.
            – Afshin
            yesterday










          • Myclass2 m2 = Myclass; is accepted by VS 2017 if the conformance mode (/permissive-) is not activated.
            – Evg
            yesterday










          • @Evg I didn't know that. I will add it too.
            – Afshin
            yesterday

















          up vote
          6
          down vote













          The line



           Myclass2 m2 = Myclass;


          means copy-initialization. Citing cppreference.com:




          If T is a class type, and the cv-unqualified version of the type of other is not T or derived from T [...], user-defined conversion sequences that can convert from the type of other to T [...] are examined and the best one is selected through overload resolution.




          Citing further:




          A user-defined conversion consists of zero or one non-explicit single-argument constructor or non-explicit conversion function call.




          So, Myclass2 m2 = Myclass; is not acceptable because it would involve two user-defined conversions.




          Now let's take a look at



          Myclass2 m2 Myclass;


          suggested in Afshin's answer. This is a direct-initialization. The rules are different:




          The constructors of T are examined and the best match is selected by overload resolution. The constructor is then called to initialize the object.




          The constructor of Myclass2 accepts Myclass1, and you need one user-defined conversion to get Myclass1 from Myclass. Hence, it compiles.




          Note that in VS copy-initilization is treated like direct-initilization if conformance mode (/premissive-) is not activated (by default). So, VS accepts Myclass2 m2 = Myclass; treating it as direct-initilization. See this document for examples.






          share|improve this answer





























            up vote
            2
            down vote













            The other answers are burying the lede: The code you’ve written is indeed invalid. MSVC accepts it by default, but MSVC is wrong to do so. You can force MSVC to be stricter by using the command line switch /permissive-. (You should use that switch.)



            Other compilers (GCC, clang), reject it.



            All compilers accept the code once you change the copy initialisation to direct initialisation as shown in the other answers.






            share|improve this answer




















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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              10
              down vote



              accepted











              compiler doesn't execute consecutive user defined conversion




              Your teacher is right. In your code sample it means Myclass cannot be converted to Myclass1 when you assign in:



              Myclass2 m2 = Myclass;


              Because constructor expects Myclass1 when creating Myclass2, and compiler cannot consecutively convert Myclass to Myclass1 and then use it for creating Myclass2. But if you have following line:



              Myclass1 m2 = Myclass;


              It will work, because constructor of Myclass1 takes Myclass as argument.



              Update:



              You may ask why this works:



              Myclass2 m2 Myclass;


              Because in this case, constructor is called and conversion can be done implicitly unless you declare Myclass1 as explicit which will fail code compilation (Thanks Fureeish for reminder), but in:



              Myclass2 m2 = Myclass;


              is like calling copy-constructor which needs reference. so if you write it like this, it will work:



              Myclass2 m2 = Myclass1(Myclass);


              As EVG mentioned, Myclass2 m2 = Myclass; is accepted by VS 2017 if the conformance mode (/permissive-) is not activated.






              share|improve this answer


















              • 2




                Worth noting that OP mentioned explicit. In your first example in your edit, there is a temporary Myclass1 object created implicitely from Myclass instance (that's why there is a Myclass2 constructor invoked, which requires Myclass1 object). If you make Myclass1's constructor explicit, it will fail
                – Fureeish
                yesterday











              • @Fureeish I will add your comment.
                – Afshin
                yesterday










              • Myclass2 m2 = Myclass; is accepted by VS 2017 if the conformance mode (/permissive-) is not activated.
                – Evg
                yesterday










              • @Evg I didn't know that. I will add it too.
                – Afshin
                yesterday














              up vote
              10
              down vote



              accepted











              compiler doesn't execute consecutive user defined conversion




              Your teacher is right. In your code sample it means Myclass cannot be converted to Myclass1 when you assign in:



              Myclass2 m2 = Myclass;


              Because constructor expects Myclass1 when creating Myclass2, and compiler cannot consecutively convert Myclass to Myclass1 and then use it for creating Myclass2. But if you have following line:



              Myclass1 m2 = Myclass;


              It will work, because constructor of Myclass1 takes Myclass as argument.



              Update:



              You may ask why this works:



              Myclass2 m2 Myclass;


              Because in this case, constructor is called and conversion can be done implicitly unless you declare Myclass1 as explicit which will fail code compilation (Thanks Fureeish for reminder), but in:



              Myclass2 m2 = Myclass;


              is like calling copy-constructor which needs reference. so if you write it like this, it will work:



              Myclass2 m2 = Myclass1(Myclass);


              As EVG mentioned, Myclass2 m2 = Myclass; is accepted by VS 2017 if the conformance mode (/permissive-) is not activated.






              share|improve this answer


















              • 2




                Worth noting that OP mentioned explicit. In your first example in your edit, there is a temporary Myclass1 object created implicitely from Myclass instance (that's why there is a Myclass2 constructor invoked, which requires Myclass1 object). If you make Myclass1's constructor explicit, it will fail
                – Fureeish
                yesterday











              • @Fureeish I will add your comment.
                – Afshin
                yesterday










              • Myclass2 m2 = Myclass; is accepted by VS 2017 if the conformance mode (/permissive-) is not activated.
                – Evg
                yesterday










              • @Evg I didn't know that. I will add it too.
                – Afshin
                yesterday












              up vote
              10
              down vote



              accepted







              up vote
              10
              down vote



              accepted







              compiler doesn't execute consecutive user defined conversion




              Your teacher is right. In your code sample it means Myclass cannot be converted to Myclass1 when you assign in:



              Myclass2 m2 = Myclass;


              Because constructor expects Myclass1 when creating Myclass2, and compiler cannot consecutively convert Myclass to Myclass1 and then use it for creating Myclass2. But if you have following line:



              Myclass1 m2 = Myclass;


              It will work, because constructor of Myclass1 takes Myclass as argument.



              Update:



              You may ask why this works:



              Myclass2 m2 Myclass;


              Because in this case, constructor is called and conversion can be done implicitly unless you declare Myclass1 as explicit which will fail code compilation (Thanks Fureeish for reminder), but in:



              Myclass2 m2 = Myclass;


              is like calling copy-constructor which needs reference. so if you write it like this, it will work:



              Myclass2 m2 = Myclass1(Myclass);


              As EVG mentioned, Myclass2 m2 = Myclass; is accepted by VS 2017 if the conformance mode (/permissive-) is not activated.






              share|improve this answer















              compiler doesn't execute consecutive user defined conversion




              Your teacher is right. In your code sample it means Myclass cannot be converted to Myclass1 when you assign in:



              Myclass2 m2 = Myclass;


              Because constructor expects Myclass1 when creating Myclass2, and compiler cannot consecutively convert Myclass to Myclass1 and then use it for creating Myclass2. But if you have following line:



              Myclass1 m2 = Myclass;


              It will work, because constructor of Myclass1 takes Myclass as argument.



              Update:



              You may ask why this works:



              Myclass2 m2 Myclass;


              Because in this case, constructor is called and conversion can be done implicitly unless you declare Myclass1 as explicit which will fail code compilation (Thanks Fureeish for reminder), but in:



              Myclass2 m2 = Myclass;


              is like calling copy-constructor which needs reference. so if you write it like this, it will work:



              Myclass2 m2 = Myclass1(Myclass);


              As EVG mentioned, Myclass2 m2 = Myclass; is accepted by VS 2017 if the conformance mode (/permissive-) is not activated.







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited yesterday

























              answered yesterday









              Afshin

              2,5251623




              2,5251623







              • 2




                Worth noting that OP mentioned explicit. In your first example in your edit, there is a temporary Myclass1 object created implicitely from Myclass instance (that's why there is a Myclass2 constructor invoked, which requires Myclass1 object). If you make Myclass1's constructor explicit, it will fail
                – Fureeish
                yesterday











              • @Fureeish I will add your comment.
                – Afshin
                yesterday










              • Myclass2 m2 = Myclass; is accepted by VS 2017 if the conformance mode (/permissive-) is not activated.
                – Evg
                yesterday










              • @Evg I didn't know that. I will add it too.
                – Afshin
                yesterday












              • 2




                Worth noting that OP mentioned explicit. In your first example in your edit, there is a temporary Myclass1 object created implicitely from Myclass instance (that's why there is a Myclass2 constructor invoked, which requires Myclass1 object). If you make Myclass1's constructor explicit, it will fail
                – Fureeish
                yesterday











              • @Fureeish I will add your comment.
                – Afshin
                yesterday










              • Myclass2 m2 = Myclass; is accepted by VS 2017 if the conformance mode (/permissive-) is not activated.
                – Evg
                yesterday










              • @Evg I didn't know that. I will add it too.
                – Afshin
                yesterday







              2




              2




              Worth noting that OP mentioned explicit. In your first example in your edit, there is a temporary Myclass1 object created implicitely from Myclass instance (that's why there is a Myclass2 constructor invoked, which requires Myclass1 object). If you make Myclass1's constructor explicit, it will fail
              – Fureeish
              yesterday





              Worth noting that OP mentioned explicit. In your first example in your edit, there is a temporary Myclass1 object created implicitely from Myclass instance (that's why there is a Myclass2 constructor invoked, which requires Myclass1 object). If you make Myclass1's constructor explicit, it will fail
              – Fureeish
              yesterday













              @Fureeish I will add your comment.
              – Afshin
              yesterday




              @Fureeish I will add your comment.
              – Afshin
              yesterday












              Myclass2 m2 = Myclass; is accepted by VS 2017 if the conformance mode (/permissive-) is not activated.
              – Evg
              yesterday




              Myclass2 m2 = Myclass; is accepted by VS 2017 if the conformance mode (/permissive-) is not activated.
              – Evg
              yesterday












              @Evg I didn't know that. I will add it too.
              – Afshin
              yesterday




              @Evg I didn't know that. I will add it too.
              – Afshin
              yesterday












              up vote
              6
              down vote













              The line



               Myclass2 m2 = Myclass;


              means copy-initialization. Citing cppreference.com:




              If T is a class type, and the cv-unqualified version of the type of other is not T or derived from T [...], user-defined conversion sequences that can convert from the type of other to T [...] are examined and the best one is selected through overload resolution.




              Citing further:




              A user-defined conversion consists of zero or one non-explicit single-argument constructor or non-explicit conversion function call.




              So, Myclass2 m2 = Myclass; is not acceptable because it would involve two user-defined conversions.




              Now let's take a look at



              Myclass2 m2 Myclass;


              suggested in Afshin's answer. This is a direct-initialization. The rules are different:




              The constructors of T are examined and the best match is selected by overload resolution. The constructor is then called to initialize the object.




              The constructor of Myclass2 accepts Myclass1, and you need one user-defined conversion to get Myclass1 from Myclass. Hence, it compiles.




              Note that in VS copy-initilization is treated like direct-initilization if conformance mode (/premissive-) is not activated (by default). So, VS accepts Myclass2 m2 = Myclass; treating it as direct-initilization. See this document for examples.






              share|improve this answer


























                up vote
                6
                down vote













                The line



                 Myclass2 m2 = Myclass;


                means copy-initialization. Citing cppreference.com:




                If T is a class type, and the cv-unqualified version of the type of other is not T or derived from T [...], user-defined conversion sequences that can convert from the type of other to T [...] are examined and the best one is selected through overload resolution.




                Citing further:




                A user-defined conversion consists of zero or one non-explicit single-argument constructor or non-explicit conversion function call.




                So, Myclass2 m2 = Myclass; is not acceptable because it would involve two user-defined conversions.




                Now let's take a look at



                Myclass2 m2 Myclass;


                suggested in Afshin's answer. This is a direct-initialization. The rules are different:




                The constructors of T are examined and the best match is selected by overload resolution. The constructor is then called to initialize the object.




                The constructor of Myclass2 accepts Myclass1, and you need one user-defined conversion to get Myclass1 from Myclass. Hence, it compiles.




                Note that in VS copy-initilization is treated like direct-initilization if conformance mode (/premissive-) is not activated (by default). So, VS accepts Myclass2 m2 = Myclass; treating it as direct-initilization. See this document for examples.






                share|improve this answer
























                  up vote
                  6
                  down vote










                  up vote
                  6
                  down vote









                  The line



                   Myclass2 m2 = Myclass;


                  means copy-initialization. Citing cppreference.com:




                  If T is a class type, and the cv-unqualified version of the type of other is not T or derived from T [...], user-defined conversion sequences that can convert from the type of other to T [...] are examined and the best one is selected through overload resolution.




                  Citing further:




                  A user-defined conversion consists of zero or one non-explicit single-argument constructor or non-explicit conversion function call.




                  So, Myclass2 m2 = Myclass; is not acceptable because it would involve two user-defined conversions.




                  Now let's take a look at



                  Myclass2 m2 Myclass;


                  suggested in Afshin's answer. This is a direct-initialization. The rules are different:




                  The constructors of T are examined and the best match is selected by overload resolution. The constructor is then called to initialize the object.




                  The constructor of Myclass2 accepts Myclass1, and you need one user-defined conversion to get Myclass1 from Myclass. Hence, it compiles.




                  Note that in VS copy-initilization is treated like direct-initilization if conformance mode (/premissive-) is not activated (by default). So, VS accepts Myclass2 m2 = Myclass; treating it as direct-initilization. See this document for examples.






                  share|improve this answer














                  The line



                   Myclass2 m2 = Myclass;


                  means copy-initialization. Citing cppreference.com:




                  If T is a class type, and the cv-unqualified version of the type of other is not T or derived from T [...], user-defined conversion sequences that can convert from the type of other to T [...] are examined and the best one is selected through overload resolution.




                  Citing further:




                  A user-defined conversion consists of zero or one non-explicit single-argument constructor or non-explicit conversion function call.




                  So, Myclass2 m2 = Myclass; is not acceptable because it would involve two user-defined conversions.




                  Now let's take a look at



                  Myclass2 m2 Myclass;


                  suggested in Afshin's answer. This is a direct-initialization. The rules are different:




                  The constructors of T are examined and the best match is selected by overload resolution. The constructor is then called to initialize the object.




                  The constructor of Myclass2 accepts Myclass1, and you need one user-defined conversion to get Myclass1 from Myclass. Hence, it compiles.




                  Note that in VS copy-initilization is treated like direct-initilization if conformance mode (/premissive-) is not activated (by default). So, VS accepts Myclass2 m2 = Myclass; treating it as direct-initilization. See this document for examples.







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited yesterday

























                  answered yesterday









                  Evg

                  3,69221334




                  3,69221334




















                      up vote
                      2
                      down vote













                      The other answers are burying the lede: The code you’ve written is indeed invalid. MSVC accepts it by default, but MSVC is wrong to do so. You can force MSVC to be stricter by using the command line switch /permissive-. (You should use that switch.)



                      Other compilers (GCC, clang), reject it.



                      All compilers accept the code once you change the copy initialisation to direct initialisation as shown in the other answers.






                      share|improve this answer
























                        up vote
                        2
                        down vote













                        The other answers are burying the lede: The code you’ve written is indeed invalid. MSVC accepts it by default, but MSVC is wrong to do so. You can force MSVC to be stricter by using the command line switch /permissive-. (You should use that switch.)



                        Other compilers (GCC, clang), reject it.



                        All compilers accept the code once you change the copy initialisation to direct initialisation as shown in the other answers.






                        share|improve this answer






















                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          The other answers are burying the lede: The code you’ve written is indeed invalid. MSVC accepts it by default, but MSVC is wrong to do so. You can force MSVC to be stricter by using the command line switch /permissive-. (You should use that switch.)



                          Other compilers (GCC, clang), reject it.



                          All compilers accept the code once you change the copy initialisation to direct initialisation as shown in the other answers.






                          share|improve this answer












                          The other answers are burying the lede: The code you’ve written is indeed invalid. MSVC accepts it by default, but MSVC is wrong to do so. You can force MSVC to be stricter by using the command line switch /permissive-. (You should use that switch.)



                          Other compilers (GCC, clang), reject it.



                          All compilers accept the code once you change the copy initialisation to direct initialisation as shown in the other answers.







                          share|improve this answer












                          share|improve this answer



                          share|improve this answer










                          answered yesterday









                          Konrad Rudolph

                          390k1017671020




                          390k1017671020




















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