How to return an fstream (C++0x)

Question

I think I'll get right into it and start with the code:

#include <iostream>
#include <fstream>
#include <string>

class test : public std::ofstream
{
    public:
        test(const std::string& filename) { this->open(gen_filename(filename)); };
        test(const test&) = delete;
        //test(test&& old) = default; // Didn't compile
        test(test&& old) {};
    private:
        std::string gen_filename(const std::string& filename) 
        { return filename + ".tmp"; }
};

int main()
{
    auto os = test("testfile");
    os << "Test1\n";
    os << "Test2\n";
}

Basically, I need to return an ofstream. Of course you can't copy an ofstream, so I fiddled around with the code in the class test, and I got the above to compile and work as you would expect (on gcc 4.5).

But I have a bad feeling this is just due to my compiler doing "Return Value Optimization" (RTO) on "auto os = test()". Indeed, if modify to the following:

int main()
{
    auto os = test("testfile");
    os << "Test1\n";
    auto os2 = std::move(os);
    os2 << "Test2\n";
}

I no longer get both Test1 and Test2 in the output.

The thing is, the class "test" isn't copyable, so there's no chance of the ofstream being duplicated. I just want to be able to return it from a function. And I seem to be able to do that with GCC.

I'd rather not have dereference smart pointers to a heap allocated ofstream, or reopen the file, as it currently works without doing those things. I just have a feeling I'm being a little "non-standard" in my approach, so a standard way of doing what I've described would be great.

Answer 1

I'm going to answer my own question here:

In the GCC C++0x Library Features page, have a look at item 27.9, which reads:

27.9 - File-based streams - Partial - Missing move and swap operations

I guess that's probably the issue I'm having with gcc.

Answer 2

The problem is with this:

test(test&& old) {};

This lets you construct a new test from an rvalue test, yes, but it says nothing about your base, which is simply being default constructed (no open file). What you want is this:

test(test&& old) : std::ofstream(std::move(old)) {};

Which will move the stream from old into the base.

Answer 3

Does the caller need to know that you are returning an ofstream, or would it make more sense to return a streambuf, and let the caller wrap it inside a stream?