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
    32
    down vote

    favorite
    4












    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
      32
      down vote

      favorite
      4









      up vote
      32
      down vote

      favorite
      4






      4





      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 Nov 15 at 12:39









      DevSolar

      47.3k1294164




      47.3k1294164
























          1 Answer
          1






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          oldest

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          up vote
          40
          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



















          • 3




            What exactly does UB stand for?
            – John
            Nov 15 at 20:48










          • see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
            – Julien Rousé
            Nov 15 at 20:51













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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
          40
          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



















          • 3




            What exactly does UB stand for?
            – John
            Nov 15 at 20:48










          • see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
            – Julien Rousé
            Nov 15 at 20:51

















          up vote
          40
          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



















          • 3




            What exactly does UB stand for?
            – John
            Nov 15 at 20:48










          • see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
            – Julien Rousé
            Nov 15 at 20:51















          up vote
          40
          down vote



          accepted







          up vote
          40
          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 Nov 15 at 15:41









          MSalters

          133k8115267




          133k8115267










          answered Nov 15 at 12:52









          rafix07

          6,0031613




          6,0031613








          • 3




            What exactly does UB stand for?
            – John
            Nov 15 at 20:48










          • see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
            – Julien Rousé
            Nov 15 at 20:51
















          • 3




            What exactly does UB stand for?
            – John
            Nov 15 at 20:48










          • see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
            – Julien Rousé
            Nov 15 at 20:51










          3




          3




          What exactly does UB stand for?
          – John
          Nov 15 at 20:48




          What exactly does UB stand for?
          – John
          Nov 15 at 20:48












          see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
          – Julien Rousé
          Nov 15 at 20:51






          see stackoverflow.com/a/2766749/3729797 for UB, Undefined Behaviour
          – Julien Rousé
          Nov 15 at 20:51




















           

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