Why doesn't make_optional work for file streams?

Question

I'm trying out the C++17 optional-type, and thought a fitting place to use it would be a function that attempts to open a file and maybe returns the opened file. The function I wrote looks like this:

std::optional<std::fstream> openFile(std::string path) { std::fstream file; file.open(path); if (!file.is_open()) { std::cerr << "couldn't open file" << path << std::endl; return {}; } else { return std::make_optional(file); // results in compilation error } }

But when I try to compile this with g++ with -std=c++17 as one of the arguments I get a big wall of template compilation error messages, starting with:

In file included from read_file.cpp:3:0: /usr/include/c++/7/optional: In instantiation of ‘constexpr std::optional<typename std::decay<_Tp>::type> std::make_optional(_Tp&&) [with _Tp = std::basic_fstream<char>&; typename std::decay<_Tp>::type = std::basic_fstream<char>]’: read_file.cpp:16:39: required from here /usr/include/c++/7/optional:991:62: error: no matching function for call to ‘std::optional<std::basic_fstream<char> >::optional(<brace-enclosed initializer list>)’ { return optional<decay_t<_Tp>> { std::forward<_Tp>(__t) }; }

Why would it seem as if fstream can't be used with std::optional? Am I approaching this in the wrong way? If optional doesn't support stream-types, does that not limit where the type can be applied?

@ecatmur return std::make_optional(file); // results in compilation error is line 16. — Swordfish
– Swordfish, Commented Jul 4, 2020 at 21:30

Bert · Accepted Answer · 2020-07-04 21:38:31Z

10

Your code will try and copy the stream, when you pass it to make_optional. Streams can't be copied, therefore, you need to move it, i.e.,

return std::make_optional(std::move(file));

or simply

return file;

(Depending on the age of the compiler, the latter might not work.)

edited Jul 4, 2020 at 21:38

answered Jul 4, 2020 at 21:28

Bert

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5 Comments

Swordfish Over a year ago

just return file ^^

Bert Over a year ago

Indeed, I'll add it to the answer, but it hides what's actually happening a bit.

Swordfish Over a year ago

i prefer the hiding and appreciate the "it just works". ymmv.

Bert Over a year ago

I think the long form serves as education, the short form as the form for actual code.

Rasmus Källqvist Over a year ago

I'm using g++ (Ubuntu 7.5.0-3ubuntu1~18.04) 7.5.0, and just returning the file results in a compilation error because it apparently doesn't implicitly convert the stream into an optional. Explicitly moving the stream into the make_optional does work however! Thank you.

NathanOliver · Accepted Answer · 2020-07-04 21:29:47Z

std::make_optional calls the optional constructor with the form of

template < class U = value_type > constexpr optional( U&& value );

and that constructor behaves as doing

T optional_data = std::forward<U>(value)

which is going to make a copy since you passed an lvalue. Streams are not copyable, so you get an error. You have to move the stream into the optional to get it to work correctly.

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