Does move assign an std::fstream close the original stream

Question

Since c++11, we can move assign one std::fstream object to another, but I'm unable to find documentation that states, what happens if the fstream object is already associated with a file (is_open()==true).

So my question is whether in the following code, File1.txt will properly be closed or if I have to close it manually. And if I have to do it manually, what happens if I don't?

std::fstream file("File1.txt");   
file = std::fstream("File2.txt"); //will this implicitly call file.close()?             

Answer 1

Move-assignment of an fstream object will result in move-assignment of its associated filebuf. The documentation for that makes it pretty clear that the old file is closed first (as-if file.rdbuf()->close() not file.close()):

basic_filebuf& operator=(basic_filebuf&& rhs);

1. Effects: Calls this->close() then move assigns from rhs. After the move assignment *this has the observable state it would have had if it had been move constructed from rhs.
2. Returns: *this.

basic_fstream& operator=(basic_fstream&& rhs);

1. Effects: Move assigns the base and members of *this from the base and corresponding members of rhs.
2. Returns: *this.

(This is the wording from draft n4527, unchanged since at least n3485)

Answer 2

Since actual file-related machinery is 'hidden' inside corresponding buffers (streams really mostly provide formatting IO), you should be looking at std::basic_filebuf documentation:

First calls close() to close the associated file, then moves the contents of rhs into *this: the put and get buffers, the associated file, the locale, the openmode, the is_open flag, and any other state. After the move, rhs is not associated with a file and rhs.is_open() == false.

Copy from http://en.cppreference.com/w/cpp/io/basic_filebuf/operator%3D