Construct container with initializer list of iterators
c++, c++11, iterator, uniform-initialization
Solution
From §13.3.2/1 ([over.match.list])
When objects of non-aggregate class type `T` are list-initialized (8.5.4), overload resolution selects the constructor in two phases:
— Initially, the candidate functions are the initializer-list constructors (8.5.4) of the class `T` and the argument list consists of the initializer list as a single argument.
— If no viable initializer-list constructor is found, overload resolution is performed again, where the candidate functions are all the constructors of the class `T` and the argument list consists of the elements of the initializer list.
In your case the initializer list constructor is deemed non-viable (because `std::istream_iterator<std::string>` is not convertible to `std::string`), and the second condition applies. This results in the constructor taking 2 iterators to be selected.
Problem
It's possible to construct a vector with an iterator range, like this: ``` std::vector<std::string> vec(std::istream_iterator<std::string>{std::cin}, std::istream_iterator<std::string>{}); ``` But I can also compile and run code using C++11 uniform initialization syntax (note the bracers), like this: ``` std::vector<std::string> vec{std::istream_iterator<std::string>{std::cin}, std::istream_iterator<std::string>{}}; ``` What's really going on here? I know that a constructor taking an initializer list gets priority over other forms of construction. Shouldn't the compiler resolve to the constructor taking an initializer list containing 2 elements of `std::istream_iterator`? This should be an error as a `std::istream_iterator` can't be converted to the vectors value type `std::string`, right?