Are there 2 times initialization when there is a constructor with default argument [duplicate]











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  • C++ Initialising fields directly vs initialisation list in default constructor

    3 answers




My question is about how the member data with initializer get initialized where there is also a default argument in the constructor.



class InputPlay {

public:
InputPlay(std::string s = "test" ) : _s(s) { };

private:
std::string _s = "default";
};


Question:



Is there are going to be 2 times initialization for the variable _s when the construct is called? aka the _s will be initialized by the string literal default and then replace by the default argument "test" in the constructor?










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Nov 20 at 16:35


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 3




    Off-topic: As you do provide the argument by value (so a new object is created anyway), you might avoid copying this object into _s in favour of moving it: : _s(std::move(s)).
    – Aconcagua
    Nov 20 at 15:55










  • @Aconcagua very nice off-topic comments!
    – SLN
    Nov 20 at 16:24















up vote
7
down vote

favorite













This question already has an answer here:




  • C++ Initialising fields directly vs initialisation list in default constructor

    3 answers




My question is about how the member data with initializer get initialized where there is also a default argument in the constructor.



class InputPlay {

public:
InputPlay(std::string s = "test" ) : _s(s) { };

private:
std::string _s = "default";
};


Question:



Is there are going to be 2 times initialization for the variable _s when the construct is called? aka the _s will be initialized by the string literal default and then replace by the default argument "test" in the constructor?










share|improve this question















marked as duplicate by StoryTeller c++
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Nov 20 at 16:35


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 3




    Off-topic: As you do provide the argument by value (so a new object is created anyway), you might avoid copying this object into _s in favour of moving it: : _s(std::move(s)).
    – Aconcagua
    Nov 20 at 15:55










  • @Aconcagua very nice off-topic comments!
    – SLN
    Nov 20 at 16:24













up vote
7
down vote

favorite









up vote
7
down vote

favorite












This question already has an answer here:




  • C++ Initialising fields directly vs initialisation list in default constructor

    3 answers




My question is about how the member data with initializer get initialized where there is also a default argument in the constructor.



class InputPlay {

public:
InputPlay(std::string s = "test" ) : _s(s) { };

private:
std::string _s = "default";
};


Question:



Is there are going to be 2 times initialization for the variable _s when the construct is called? aka the _s will be initialized by the string literal default and then replace by the default argument "test" in the constructor?










share|improve this question
















This question already has an answer here:




  • C++ Initialising fields directly vs initialisation list in default constructor

    3 answers




My question is about how the member data with initializer get initialized where there is also a default argument in the constructor.



class InputPlay {

public:
InputPlay(std::string s = "test" ) : _s(s) { };

private:
std::string _s = "default";
};


Question:



Is there are going to be 2 times initialization for the variable _s when the construct is called? aka the _s will be initialized by the string literal default and then replace by the default argument "test" in the constructor?





This question already has an answer here:




  • C++ Initialising fields directly vs initialisation list in default constructor

    3 answers








c++ constructor initialization






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edited Nov 21 at 5:36









Cœur

17k9102140




17k9102140










asked Nov 20 at 15:46









SLN

1,4911429




1,4911429




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Nov 20 at 16:35


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






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Nov 20 at 16:35


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 3




    Off-topic: As you do provide the argument by value (so a new object is created anyway), you might avoid copying this object into _s in favour of moving it: : _s(std::move(s)).
    – Aconcagua
    Nov 20 at 15:55










  • @Aconcagua very nice off-topic comments!
    – SLN
    Nov 20 at 16:24














  • 3




    Off-topic: As you do provide the argument by value (so a new object is created anyway), you might avoid copying this object into _s in favour of moving it: : _s(std::move(s)).
    – Aconcagua
    Nov 20 at 15:55










  • @Aconcagua very nice off-topic comments!
    – SLN
    Nov 20 at 16:24








3




3




Off-topic: As you do provide the argument by value (so a new object is created anyway), you might avoid copying this object into _s in favour of moving it: : _s(std::move(s)).
– Aconcagua
Nov 20 at 15:55




Off-topic: As you do provide the argument by value (so a new object is created anyway), you might avoid copying this object into _s in favour of moving it: : _s(std::move(s)).
– Aconcagua
Nov 20 at 15:55












@Aconcagua very nice off-topic comments!
– SLN
Nov 20 at 16:24




@Aconcagua very nice off-topic comments!
– SLN
Nov 20 at 16:24












3 Answers
3






active

oldest

votes

















up vote
9
down vote



accepted










No, _s will only be initialized once. The in class initialization is syntactic sugar for synthesizing a member initializer. If you provide your own member initializer then the compiler will use that instead of synthesizing one for you from the in class initialization.






share|improve this answer




























    up vote
    6
    down vote













    No. For non-static data member, when both default member initializer and member initializer list are provided, the default member initializer will be ignored. That means _s will be initialized by the argument s of the constructor directly.




    If a member has a default member initializer and also appears in the member initialization list in a constructor, the default member initializer is ignored.







    share|improve this answer




























      up vote
      5
      down vote













      = "default"; is a default member initializer. It's the initializer to use for _s if you don't provide one. But since you do provide one (_s(s)) the default initializer never comes into player and _s is simply initialized to s.



      From cppreference.com :




      If a non-static data member has an default member initializer and also appears in a member initializer list, then member initializer list is executed and the default member initializer is ignored:




      Note that by definition something can only ever be initialized once. If something looks like it's initialized twice, it's usually being initialized and then assigned another value.






      share|improve this answer




























        3 Answers
        3






        active

        oldest

        votes








        3 Answers
        3






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes








        up vote
        9
        down vote



        accepted










        No, _s will only be initialized once. The in class initialization is syntactic sugar for synthesizing a member initializer. If you provide your own member initializer then the compiler will use that instead of synthesizing one for you from the in class initialization.






        share|improve this answer

























          up vote
          9
          down vote



          accepted










          No, _s will only be initialized once. The in class initialization is syntactic sugar for synthesizing a member initializer. If you provide your own member initializer then the compiler will use that instead of synthesizing one for you from the in class initialization.






          share|improve this answer























            up vote
            9
            down vote



            accepted







            up vote
            9
            down vote



            accepted






            No, _s will only be initialized once. The in class initialization is syntactic sugar for synthesizing a member initializer. If you provide your own member initializer then the compiler will use that instead of synthesizing one for you from the in class initialization.






            share|improve this answer












            No, _s will only be initialized once. The in class initialization is syntactic sugar for synthesizing a member initializer. If you provide your own member initializer then the compiler will use that instead of synthesizing one for you from the in class initialization.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 20 at 15:50









            NathanOliver

            82.7k15112172




            82.7k15112172
























                up vote
                6
                down vote













                No. For non-static data member, when both default member initializer and member initializer list are provided, the default member initializer will be ignored. That means _s will be initialized by the argument s of the constructor directly.




                If a member has a default member initializer and also appears in the member initialization list in a constructor, the default member initializer is ignored.







                share|improve this answer

























                  up vote
                  6
                  down vote













                  No. For non-static data member, when both default member initializer and member initializer list are provided, the default member initializer will be ignored. That means _s will be initialized by the argument s of the constructor directly.




                  If a member has a default member initializer and also appears in the member initialization list in a constructor, the default member initializer is ignored.







                  share|improve this answer























                    up vote
                    6
                    down vote










                    up vote
                    6
                    down vote









                    No. For non-static data member, when both default member initializer and member initializer list are provided, the default member initializer will be ignored. That means _s will be initialized by the argument s of the constructor directly.




                    If a member has a default member initializer and also appears in the member initialization list in a constructor, the default member initializer is ignored.







                    share|improve this answer












                    No. For non-static data member, when both default member initializer and member initializer list are provided, the default member initializer will be ignored. That means _s will be initialized by the argument s of the constructor directly.




                    If a member has a default member initializer and also appears in the member initialization list in a constructor, the default member initializer is ignored.








                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered Nov 20 at 15:50









                    songyuanyao

                    88.4k11170232




                    88.4k11170232






















                        up vote
                        5
                        down vote













                        = "default"; is a default member initializer. It's the initializer to use for _s if you don't provide one. But since you do provide one (_s(s)) the default initializer never comes into player and _s is simply initialized to s.



                        From cppreference.com :




                        If a non-static data member has an default member initializer and also appears in a member initializer list, then member initializer list is executed and the default member initializer is ignored:




                        Note that by definition something can only ever be initialized once. If something looks like it's initialized twice, it's usually being initialized and then assigned another value.






                        share|improve this answer

























                          up vote
                          5
                          down vote













                          = "default"; is a default member initializer. It's the initializer to use for _s if you don't provide one. But since you do provide one (_s(s)) the default initializer never comes into player and _s is simply initialized to s.



                          From cppreference.com :




                          If a non-static data member has an default member initializer and also appears in a member initializer list, then member initializer list is executed and the default member initializer is ignored:




                          Note that by definition something can only ever be initialized once. If something looks like it's initialized twice, it's usually being initialized and then assigned another value.






                          share|improve this answer























                            up vote
                            5
                            down vote










                            up vote
                            5
                            down vote









                            = "default"; is a default member initializer. It's the initializer to use for _s if you don't provide one. But since you do provide one (_s(s)) the default initializer never comes into player and _s is simply initialized to s.



                            From cppreference.com :




                            If a non-static data member has an default member initializer and also appears in a member initializer list, then member initializer list is executed and the default member initializer is ignored:




                            Note that by definition something can only ever be initialized once. If something looks like it's initialized twice, it's usually being initialized and then assigned another value.






                            share|improve this answer












                            = "default"; is a default member initializer. It's the initializer to use for _s if you don't provide one. But since you do provide one (_s(s)) the default initializer never comes into player and _s is simply initialized to s.



                            From cppreference.com :




                            If a non-static data member has an default member initializer and also appears in a member initializer list, then member initializer list is executed and the default member initializer is ignored:




                            Note that by definition something can only ever be initialized once. If something looks like it's initialized twice, it's usually being initialized and then assigned another value.







                            share|improve this answer












                            share|improve this answer



                            share|improve this answer










                            answered Nov 20 at 15:54









                            François Andrieux

                            14.8k32345




                            14.8k32345















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