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Why would a new[] expression ever invoke a destructor?


About constructors/destructors and new/delete operators in C++ for custom objectsDoes delete call the destructor?Overloading operator delete in a base classWhen is %destructor invoked in BISON?Why should C++ programmers minimize use of 'new'?Why is ::operator new[] necessary when ::operator new is enough?Why am I permitted to declare an object with a deleted destructor?Does delete[] deallocate memory in one shot after invoking destructors?Why are non-placement `new` and `delete` built into the language and not just regular functions?Why deallocation function is not called when object constructor throw in a new expression?













8















From the C++17 standard (draft here), [expr.new]:




If the new-expression creates an object or an array of objects of class type, access and ambiguity control are done for the allocation function, the deallocation function, and the constructor. If the new-expression creates an array of objects of class type, the destructor is potentially invoked.




Why would new[] invoke a destructor? It's new, after all. It isn't delete.










share|improve this question




























    8















    From the C++17 standard (draft here), [expr.new]:




    If the new-expression creates an object or an array of objects of class type, access and ambiguity control are done for the allocation function, the deallocation function, and the constructor. If the new-expression creates an array of objects of class type, the destructor is potentially invoked.




    Why would new[] invoke a destructor? It's new, after all. It isn't delete.










    share|improve this question


























      8












      8








      8








      From the C++17 standard (draft here), [expr.new]:




      If the new-expression creates an object or an array of objects of class type, access and ambiguity control are done for the allocation function, the deallocation function, and the constructor. If the new-expression creates an array of objects of class type, the destructor is potentially invoked.




      Why would new[] invoke a destructor? It's new, after all. It isn't delete.










      share|improve this question
















      From the C++17 standard (draft here), [expr.new]:




      If the new-expression creates an object or an array of objects of class type, access and ambiguity control are done for the allocation function, the deallocation function, and the constructor. If the new-expression creates an array of objects of class type, the destructor is potentially invoked.




      Why would new[] invoke a destructor? It's new, after all. It isn't delete.







      c++ c++17 new-operator






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      share|improve this question








      edited 1 hour ago









      Cody Gray

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      195k35382470










      asked 1 hour ago









      thbthb

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      8,66932355






















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          If construction of any object in the buffer throws an exception, the previously constructed objects must be destructed. That requires an available destructor.






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            If construction of any object in the buffer throws an exception, the previously constructed objects must be destructed. That requires an available destructor.






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              14














              If construction of any object in the buffer throws an exception, the previously constructed objects must be destructed. That requires an available destructor.






              share|improve this answer



























                14












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                14







                If construction of any object in the buffer throws an exception, the previously constructed objects must be destructed. That requires an available destructor.






                share|improve this answer















                If construction of any object in the buffer throws an exception, the previously constructed objects must be destructed. That requires an available destructor.







                share|improve this answer














                share|improve this answer



                share|improve this answer








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