Cpp constructor6/23/2023 ![]() If the pack has one or more arguments that could all be converted to an int, then this constructor is instantiated and called. The initialization list comes after the constructor signature following a colon, and before the body of the constructor. ![]() In the above code, a fold expression has been used to check whether each of the arguments in the pack could be converted to an integral type. In C++ there is a concept of constructor's initialization list, which is where you can and should call the base class' constructor and where you should also initialize the data members. A way based on concepts (C++20) is #include The answer by Dharmesh946 shows a way based on enable_if. It can be achieved via enable_if (again, refer to item 27 of the above Scott Meyer's book) or via concepts in C++20. The constructor taking the universal reference can be constrained, so that it will be called only if a list of values to be assigned to the underlying array is passed. Then between a constructor taking A& (the instantiated templated constructor) and const A& (the already present copy constructor), it will choose the instantiated constructor because it is simpler (no need to add constness to A&). ![]() It will instantiate the templated constructor to take a non-const lvalue A&. Given the presence of the templated constructor, the compiler is obliged to go through the process of overload resolution to find the best possible function. Creating a temporary and then assigning to it is too much work. This is because the variable b is being created in this statement. Also, the constructor based on the universal reference being variadic in nature is not relevant to this discussion.Īs such, A b = a will lead the compiler to call a constructor that will take a copy of A and it will not call the copy assignment operator. When the compiler does overload resolution, the constructor based on universal reference (also called forwarding reference) will be the function that will be preferred by the compiler. Refer to item 26 of "Effective modern C++" by Scott Meyers: Avoid overloading on universal references. A c(a) // also fails (templated constructor is better match)Ĭompile output: templated_constructror.cpp: In instantiation of ‘A::A(Args&. Std::cout << _PRETTY_FUNCTION_ << std::endl (Actually I can see how the compiler wouldn't be able to discern the copy constructor signature as different from the templated constructor, but seems like it should be obvious when the copy assignment operator is used) Trying to understand why having a parameter pack templated constructor for a class apparently causes both the copy constructor and the copy assignment operator to be optimized out.
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