I'm new to C++11 and I was wondering how this code works internally:
class MyClass { public: MyClass(int a, double b) { cout << "ctor()" << endl; } }; int main() { MyClass i1{4, 7}; return 0; } My understanding of the new initializer list is that it is a class std::initializer_list constructed by special syntax { .... } in the code. So how does this class instance created by {4, 7} internally get transformed to a form that fits the constructor to MyClass? Thanks.
I think this is how it happens. Extracted from: Explanation of list initialization at cppreference.com
If the previous stage does not produce a match, all constructors of T participate in overload resolution against the set of arguments that consists of the elements of the braced-init-list, with the restriction that only non-narrowing conversions are allowed. If this stage produces an explicit constructor as the best match for a copy-list-initialization, compilation fails (note, in simple copy-initialization, explicit constructors are not considered at all)
{....}can construct astd::initializer_list... or it can do other things.