问题
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?
回答1:
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.
来源:https://stackoverflow.com/questions/17979857/construct-container-with-initializer-list-of-iterators