#include <iostream> class A{ }; class B: public A{ public: B(A&& inA){ std::cout<<"constructor"<<std::endl; } }; int main(){ B whatever{A{}}; whatever=A{}; } This outputs
constructor constructor at least with C++14 standard and GCC. How is it defined that assignment operator can result in call to constructor instead of operator=? Is there a name for this property of assignment operator?
Since you meet all the conditions for generating a move-assignment operator. The move-assignment operator the compiler synthesizes for you is in the form of:
B& operator=(B&&) = default; Recall that temporaries can be bound to const lvalue references and rvalue references. By Implicit Conversion Sequences, your temporary A{} is converted to a temporary B which is used to make the move assignment. You may disable this with explicit constructors.
B->A)" conversion, and there is no such context in OP's snippet.... OP's code has a Converting Constructor[class.conv.ctor] that takes an rvalue to an object of A, namely: B(A&&). This means any rvalue object of A can be converted to B. User-defined Conversion Sequences[over.ics.user] takes into account all possible methods of implicit conversion from "the Type you passed as argument" to "the Type expected"