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8















I am writing some code and have a question.
What is the difference between auto x = vector<int>(); and vector<int> x;? Are they all the same or there's some difference with the complexity?










share|improve this question




















  • 2





    The first results in a call to a default constructor and a call to a move constructor. The second results in a call to a default constructor. Even if the compiler optimizes both to result in the same assembly, the second one is the one to go for readability.

    – DeiDei
    1 hour ago


















8















I am writing some code and have a question.
What is the difference between auto x = vector<int>(); and vector<int> x;? Are they all the same or there's some difference with the complexity?










share|improve this question




















  • 2





    The first results in a call to a default constructor and a call to a move constructor. The second results in a call to a default constructor. Even if the compiler optimizes both to result in the same assembly, the second one is the one to go for readability.

    – DeiDei
    1 hour ago














8












8








8


2






I am writing some code and have a question.
What is the difference between auto x = vector<int>(); and vector<int> x;? Are they all the same or there's some difference with the complexity?










share|improve this question
















I am writing some code and have a question.
What is the difference between auto x = vector<int>(); and vector<int> x;? Are they all the same or there's some difference with the complexity?







c++ vector stl initialization






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









songyuanyao

94.5k11182250




94.5k11182250










asked 1 hour ago









AutoratchAutoratch

555




555








  • 2





    The first results in a call to a default constructor and a call to a move constructor. The second results in a call to a default constructor. Even if the compiler optimizes both to result in the same assembly, the second one is the one to go for readability.

    – DeiDei
    1 hour ago














  • 2





    The first results in a call to a default constructor and a call to a move constructor. The second results in a call to a default constructor. Even if the compiler optimizes both to result in the same assembly, the second one is the one to go for readability.

    – DeiDei
    1 hour ago








2




2





The first results in a call to a default constructor and a call to a move constructor. The second results in a call to a default constructor. Even if the compiler optimizes both to result in the same assembly, the second one is the one to go for readability.

– DeiDei
1 hour ago





The first results in a call to a default constructor and a call to a move constructor. The second results in a call to a default constructor. Even if the compiler optimizes both to result in the same assembly, the second one is the one to go for readability.

– DeiDei
1 hour ago












1 Answer
1






active

oldest

votes


















9














They have the same effect since C++17. Both construct an object named x with type std::vector<int>, which is initialized by the default constructor of std::vector.



Precisely the 1st one is copy initialization, x is copy-initialized from a value-initialized temporary. From C++17 this kind of copy elision is guaranteed, as the result x is initialized by the default constructor of std::vector directly. Before C++17, copy elision is an optimization:




even when it takes place and the copy/move (since C++11) constructor is not called, it still must be present and accessible (as if no optimization happened at all), otherwise the program is ill-formed:




The 2nd one is default initialization, as a class type x is initialized by the default constructor of std::vector.






share|improve this answer


























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    1 Answer
    1






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    9














    They have the same effect since C++17. Both construct an object named x with type std::vector<int>, which is initialized by the default constructor of std::vector.



    Precisely the 1st one is copy initialization, x is copy-initialized from a value-initialized temporary. From C++17 this kind of copy elision is guaranteed, as the result x is initialized by the default constructor of std::vector directly. Before C++17, copy elision is an optimization:




    even when it takes place and the copy/move (since C++11) constructor is not called, it still must be present and accessible (as if no optimization happened at all), otherwise the program is ill-formed:




    The 2nd one is default initialization, as a class type x is initialized by the default constructor of std::vector.






    share|improve this answer






























      9














      They have the same effect since C++17. Both construct an object named x with type std::vector<int>, which is initialized by the default constructor of std::vector.



      Precisely the 1st one is copy initialization, x is copy-initialized from a value-initialized temporary. From C++17 this kind of copy elision is guaranteed, as the result x is initialized by the default constructor of std::vector directly. Before C++17, copy elision is an optimization:




      even when it takes place and the copy/move (since C++11) constructor is not called, it still must be present and accessible (as if no optimization happened at all), otherwise the program is ill-formed:




      The 2nd one is default initialization, as a class type x is initialized by the default constructor of std::vector.






      share|improve this answer




























        9












        9








        9







        They have the same effect since C++17. Both construct an object named x with type std::vector<int>, which is initialized by the default constructor of std::vector.



        Precisely the 1st one is copy initialization, x is copy-initialized from a value-initialized temporary. From C++17 this kind of copy elision is guaranteed, as the result x is initialized by the default constructor of std::vector directly. Before C++17, copy elision is an optimization:




        even when it takes place and the copy/move (since C++11) constructor is not called, it still must be present and accessible (as if no optimization happened at all), otherwise the program is ill-formed:




        The 2nd one is default initialization, as a class type x is initialized by the default constructor of std::vector.






        share|improve this answer















        They have the same effect since C++17. Both construct an object named x with type std::vector<int>, which is initialized by the default constructor of std::vector.



        Precisely the 1st one is copy initialization, x is copy-initialized from a value-initialized temporary. From C++17 this kind of copy elision is guaranteed, as the result x is initialized by the default constructor of std::vector directly. Before C++17, copy elision is an optimization:




        even when it takes place and the copy/move (since C++11) constructor is not called, it still must be present and accessible (as if no optimization happened at all), otherwise the program is ill-formed:




        The 2nd one is default initialization, as a class type x is initialized by the default constructor of std::vector.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 1 hour ago

























        answered 1 hour ago









        songyuanyaosongyuanyao

        94.5k11182250




        94.5k11182250
































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