Is std::next for vector O(n) or O(1)?





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11















In C++11 I use std::next because If I want to change vector to list, I dont have to change the rest of code.



For list, std::next is O(n), because I need to iterate all elements. But how is it for a vector? I have found this note on cppreference:




However, if InputIt or ForwardIt additionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.




Does vector meet these requirements? And why "Legacy"?






c++







share|improve this question




















  • 12





    "Constant" means O(1).

    – Some programmer dude
    9 hours ago






  • 1





    I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?

    – Martin Perry
    9 hours ago








  • 4





    @MartinPerry Follow the link? In the page you pull the quote from, the word "LegacyRandomAccessIterator" is a link to an explanation about it. And for vectors, iterators are (legacy) random-access iterators (as you should be able to find out in the std::vector reference).

    – Some programmer dude
    9 hours ago




















11















In C++11 I use std::next because If I want to change vector to list, I dont have to change the rest of code.



For list, std::next is O(n), because I need to iterate all elements. But how is it for a vector? I have found this note on cppreference:




However, if InputIt or ForwardIt additionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.




Does vector meet these requirements? And why "Legacy"?






c++







share|improve this question




















  • 12





    "Constant" means O(1).

    – Some programmer dude
    9 hours ago






  • 1





    I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?

    – Martin Perry
    9 hours ago








  • 4





    @MartinPerry Follow the link? In the page you pull the quote from, the word "LegacyRandomAccessIterator" is a link to an explanation about it. And for vectors, iterators are (legacy) random-access iterators (as you should be able to find out in the std::vector reference).

    – Some programmer dude
    9 hours ago
















11












11








11


2






In C++11 I use std::next because If I want to change vector to list, I dont have to change the rest of code.



For list, std::next is O(n), because I need to iterate all elements. But how is it for a vector? I have found this note on cppreference:




However, if InputIt or ForwardIt additionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.




Does vector meet these requirements? And why "Legacy"?






c++







share|improve this question
















In C++11 I use std::next because If I want to change vector to list, I dont have to change the rest of code.



For list, std::next is O(n), because I need to iterate all elements. But how is it for a vector? I have found this note on cppreference:




However, if InputIt or ForwardIt additionally meets the requirements of LegacyRandomAccessIterator, complexity is constant.




Does vector meet these requirements? And why "Legacy"?






c++




c++






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 9 hours ago









Angew

135k11261354




135k11261354










asked 9 hours ago









Martin PerryMartin Perry

5,21833267




5,21833267








  • 12





    "Constant" means O(1).

    – Some programmer dude
    9 hours ago






  • 1





    I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?

    – Martin Perry
    9 hours ago








  • 4





    @MartinPerry Follow the link? In the page you pull the quote from, the word "LegacyRandomAccessIterator" is a link to an explanation about it. And for vectors, iterators are (legacy) random-access iterators (as you should be able to find out in the std::vector reference).

    – Some programmer dude
    9 hours ago
















  • 12





    "Constant" means O(1).

    – Some programmer dude
    9 hours ago






  • 1





    I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?

    – Martin Perry
    9 hours ago








  • 4





    @MartinPerry Follow the link? In the page you pull the quote from, the word "LegacyRandomAccessIterator" is a link to an explanation about it. And for vectors, iterators are (legacy) random-access iterators (as you should be able to find out in the std::vector reference).

    – Some programmer dude
    9 hours ago










12




12





"Constant" means O(1).

– Some programmer dude
9 hours ago





"Constant" means O(1).

– Some programmer dude
9 hours ago




1




1





I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?

– Martin Perry
9 hours ago







I know that, but I dont know If it aplies to vector, since I dont understand LegacyRandomAccessIterator. Why Legacy?

– Martin Perry
9 hours ago






4




4





@MartinPerry Follow the link? In the page you pull the quote from, the word "LegacyRandomAccessIterator" is a link to an explanation about it. And for vectors, iterators are (legacy) random-access iterators (as you should be able to find out in the std::vector reference).

– Some programmer dude
9 hours ago







@MartinPerry Follow the link? In the page you pull the quote from, the word "LegacyRandomAccessIterator" is a link to an explanation about it. And for vectors, iterators are (legacy) random-access iterators (as you should be able to find out in the std::vector reference).

– Some programmer dude
9 hours ago














2 Answers
2






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14














There is a plan of adding concepts (compile time type constraints) in C++20. The new standard is supposed to contain concepts like InputIterator or RandomAccessIterator. To distinguish between the concepts and the old trait-like requirements cppreference uses LegacyRandomAccessIterator and so on for pre-concept requirements and RandomAccessIterator and so for concept requirements.



And so yes, std::vector::iterator meets requirements of LegacyRandomAccessIterator and actually will be fulfilling RandomAccessIterator concept as well. This leads straight to conclusion that std::next called on vector::iterator has complexity O(1).






share|improve this answer

































    8















    Does vector meet these requirements?




    Yes, it does:



    https://en.cppreference.com/w/cpp/container/vector



    Quote: "iterator LegacyRandomAccessIterator"




    And why "Legacy"?




    The existing iterators have been renamed "legacy" because of the upcoming C++ library feature called ranges, which is a replacement for the current approach. Ranges will have new iterators. The existing ones will still be there, thus they're called "legacy."






    share|improve this answer
























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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      14














      There is a plan of adding concepts (compile time type constraints) in C++20. The new standard is supposed to contain concepts like InputIterator or RandomAccessIterator. To distinguish between the concepts and the old trait-like requirements cppreference uses LegacyRandomAccessIterator and so on for pre-concept requirements and RandomAccessIterator and so for concept requirements.



      And so yes, std::vector::iterator meets requirements of LegacyRandomAccessIterator and actually will be fulfilling RandomAccessIterator concept as well. This leads straight to conclusion that std::next called on vector::iterator has complexity O(1).






      share|improve this answer






























        14














        There is a plan of adding concepts (compile time type constraints) in C++20. The new standard is supposed to contain concepts like InputIterator or RandomAccessIterator. To distinguish between the concepts and the old trait-like requirements cppreference uses LegacyRandomAccessIterator and so on for pre-concept requirements and RandomAccessIterator and so for concept requirements.



        And so yes, std::vector::iterator meets requirements of LegacyRandomAccessIterator and actually will be fulfilling RandomAccessIterator concept as well. This leads straight to conclusion that std::next called on vector::iterator has complexity O(1).






        share|improve this answer




























          14












          14








          14







          There is a plan of adding concepts (compile time type constraints) in C++20. The new standard is supposed to contain concepts like InputIterator or RandomAccessIterator. To distinguish between the concepts and the old trait-like requirements cppreference uses LegacyRandomAccessIterator and so on for pre-concept requirements and RandomAccessIterator and so for concept requirements.



          And so yes, std::vector::iterator meets requirements of LegacyRandomAccessIterator and actually will be fulfilling RandomAccessIterator concept as well. This leads straight to conclusion that std::next called on vector::iterator has complexity O(1).






          share|improve this answer















          There is a plan of adding concepts (compile time type constraints) in C++20. The new standard is supposed to contain concepts like InputIterator or RandomAccessIterator. To distinguish between the concepts and the old trait-like requirements cppreference uses LegacyRandomAccessIterator and so on for pre-concept requirements and RandomAccessIterator and so for concept requirements.



          And so yes, std::vector::iterator meets requirements of LegacyRandomAccessIterator and actually will be fulfilling RandomAccessIterator concept as well. This leads straight to conclusion that std::next called on vector::iterator has complexity O(1).







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 9 hours ago

























          answered 9 hours ago









          bartopbartop

          3,4151132




          3,4151132

























              8















              Does vector meet these requirements?




              Yes, it does:



              https://en.cppreference.com/w/cpp/container/vector



              Quote: "iterator LegacyRandomAccessIterator"




              And why "Legacy"?




              The existing iterators have been renamed "legacy" because of the upcoming C++ library feature called ranges, which is a replacement for the current approach. Ranges will have new iterators. The existing ones will still be there, thus they're called "legacy."






              share|improve this answer




























                8















                Does vector meet these requirements?




                Yes, it does:



                https://en.cppreference.com/w/cpp/container/vector



                Quote: "iterator LegacyRandomAccessIterator"




                And why "Legacy"?




                The existing iterators have been renamed "legacy" because of the upcoming C++ library feature called ranges, which is a replacement for the current approach. Ranges will have new iterators. The existing ones will still be there, thus they're called "legacy."






                share|improve this answer


























                  8












                  8








                  8








                  Does vector meet these requirements?




                  Yes, it does:



                  https://en.cppreference.com/w/cpp/container/vector



                  Quote: "iterator LegacyRandomAccessIterator"




                  And why "Legacy"?




                  The existing iterators have been renamed "legacy" because of the upcoming C++ library feature called ranges, which is a replacement for the current approach. Ranges will have new iterators. The existing ones will still be there, thus they're called "legacy."






                  share|improve this answer














                  Does vector meet these requirements?




                  Yes, it does:



                  https://en.cppreference.com/w/cpp/container/vector



                  Quote: "iterator LegacyRandomAccessIterator"




                  And why "Legacy"?




                  The existing iterators have been renamed "legacy" because of the upcoming C++ library feature called ranges, which is a replacement for the current approach. Ranges will have new iterators. The existing ones will still be there, thus they're called "legacy."







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 9 hours ago









                  Nikos C.Nikos C.

                  34.2k53967




                  34.2k53967






























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