CppCon 2018, Nicolai Josuttis: Why are these interpreted as iterators?

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Nicolai Josuttis' "The Nightmare of Initialization in C++" presentation at CppCon 2018 had, at one point, the following piece of code:



std::vector< std::string > v07 = "1", "2" ;


Nicolai said the following (transcription mine):




The problem is, what happens here is, we interpret these two parameters as iterators. So these are iterators, so this is the beginning of the range, and this is the end of the range, and they should refer to the same range of characters; because characters convert implicitly to strings this will compile. If you're lucky, you'll get a coredump. If not, you've got a big problem.




He lost me there. Can somebody explain what is going on here, exactly, step by step?










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    up vote
    24
    down vote

    favorite
    1












    Nicolai Josuttis' "The Nightmare of Initialization in C++" presentation at CppCon 2018 had, at one point, the following piece of code:



    std::vector< std::string > v07 = "1", "2" ;


    Nicolai said the following (transcription mine):




    The problem is, what happens here is, we interpret these two parameters as iterators. So these are iterators, so this is the beginning of the range, and this is the end of the range, and they should refer to the same range of characters; because characters convert implicitly to strings this will compile. If you're lucky, you'll get a coredump. If not, you've got a big problem.




    He lost me there. Can somebody explain what is going on here, exactly, step by step?










    share|improve this question























      up vote
      24
      down vote

      favorite
      1









      up vote
      24
      down vote

      favorite
      1






      1





      Nicolai Josuttis' "The Nightmare of Initialization in C++" presentation at CppCon 2018 had, at one point, the following piece of code:



      std::vector< std::string > v07 = "1", "2" ;


      Nicolai said the following (transcription mine):




      The problem is, what happens here is, we interpret these two parameters as iterators. So these are iterators, so this is the beginning of the range, and this is the end of the range, and they should refer to the same range of characters; because characters convert implicitly to strings this will compile. If you're lucky, you'll get a coredump. If not, you've got a big problem.




      He lost me there. Can somebody explain what is going on here, exactly, step by step?










      share|improve this question













      Nicolai Josuttis' "The Nightmare of Initialization in C++" presentation at CppCon 2018 had, at one point, the following piece of code:



      std::vector< std::string > v07 = "1", "2" ;


      Nicolai said the following (transcription mine):




      The problem is, what happens here is, we interpret these two parameters as iterators. So these are iterators, so this is the beginning of the range, and this is the end of the range, and they should refer to the same range of characters; because characters convert implicitly to strings this will compile. If you're lucky, you'll get a coredump. If not, you've got a big problem.




      He lost me there. Can somebody explain what is going on here, exactly, step by step?







      c++ initialization c++17






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      asked 2 days ago









      DevSolar

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      47.1k1293164






















          1 Answer
          1






          active

          oldest

          votes

















          up vote
          30
          down vote



          accepted










          Below code



          std::vector< std::string > v07 = "1", "2" ;


          is equivalent to



          std::string s = "1","2"; // call string(const char*, const char*)
          std::vector<std::string> v07 = s; // initializer list with one item


          the issue is with



           s="1","2";


          This calls string(const char* start, const char* end) constructor,
          but start and end must refer to the same string object. "1" and "2" are two different objects, so it leads to UB.






          share|improve this answer


















          • 1




            What exactly does UB stand for?
            – John
            2 days ago










          • see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
            – Julien Rousé
            2 days ago











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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes








          up vote
          30
          down vote



          accepted










          Below code



          std::vector< std::string > v07 = "1", "2" ;


          is equivalent to



          std::string s = "1","2"; // call string(const char*, const char*)
          std::vector<std::string> v07 = s; // initializer list with one item


          the issue is with



           s="1","2";


          This calls string(const char* start, const char* end) constructor,
          but start and end must refer to the same string object. "1" and "2" are two different objects, so it leads to UB.






          share|improve this answer


















          • 1




            What exactly does UB stand for?
            – John
            2 days ago










          • see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
            – Julien Rousé
            2 days ago















          up vote
          30
          down vote



          accepted










          Below code



          std::vector< std::string > v07 = "1", "2" ;


          is equivalent to



          std::string s = "1","2"; // call string(const char*, const char*)
          std::vector<std::string> v07 = s; // initializer list with one item


          the issue is with



           s="1","2";


          This calls string(const char* start, const char* end) constructor,
          but start and end must refer to the same string object. "1" and "2" are two different objects, so it leads to UB.






          share|improve this answer


















          • 1




            What exactly does UB stand for?
            – John
            2 days ago










          • see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
            – Julien Rousé
            2 days ago













          up vote
          30
          down vote



          accepted







          up vote
          30
          down vote



          accepted






          Below code



          std::vector< std::string > v07 = "1", "2" ;


          is equivalent to



          std::string s = "1","2"; // call string(const char*, const char*)
          std::vector<std::string> v07 = s; // initializer list with one item


          the issue is with



           s="1","2";


          This calls string(const char* start, const char* end) constructor,
          but start and end must refer to the same string object. "1" and "2" are two different objects, so it leads to UB.






          share|improve this answer














          Below code



          std::vector< std::string > v07 = "1", "2" ;


          is equivalent to



          std::string s = "1","2"; // call string(const char*, const char*)
          std::vector<std::string> v07 = s; // initializer list with one item


          the issue is with



           s="1","2";


          This calls string(const char* start, const char* end) constructor,
          but start and end must refer to the same string object. "1" and "2" are two different objects, so it leads to UB.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 2 days ago









          MSalters

          132k8115267




          132k8115267










          answered 2 days ago









          rafix07

          5,7731513




          5,7731513







          • 1




            What exactly does UB stand for?
            – John
            2 days ago










          • see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
            – Julien Rousé
            2 days ago













          • 1




            What exactly does UB stand for?
            – John
            2 days ago










          • see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
            – Julien Rousé
            2 days ago








          1




          1




          What exactly does UB stand for?
          – John
          2 days ago




          What exactly does UB stand for?
          – John
          2 days ago












          see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
          – Julien Rousé
          2 days ago





          see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
          – Julien Rousé
          2 days ago


















           

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