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I modeled a function that returns an instance of an std::expected<void, Error> - I told myself "I can use a new standard so I will design my library accordingly" - and I was very positive to have nice and concrete error handling. Now it turns out, that all the monadic operations on the std::expected only work on non-void items. Even though there is a specialization for void, the monadic operations are not available. I understand that or_else needs to return the value - but there is a specialization for void, so why should this not work?

std::expected<void, Error> fun (int);

fun (19).and_then ([]() { doSomething(); }).or_else ([] (Error e) { std::println ("Uh oh error.."); });

This yields:

/usr/include/c++/14.2.1/expected:1586:37: error: static assertion failed
 1586 |           static_assert(__expected::__is_expected<_Up>);
      |                         ~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~
/usr/include/c++/14.2.1/expected:1586:37: note: ‘std::__expected::__is_expected<void>’ evaluates to false
/usr/include/c++/14.2.1/expected:1587:25: error: ‘std::remove_cvref<void>::type’ {aka ‘void’} is not a class, struct, or union type
 1587 |           static_assert(is_same_v<typename _Up::error_type, _Er>);
      |                         ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/usr/include/c++/14.2.1/expected:1592:20: error: expression list treated as compound expression in functional cast [-fpermissive]
 1592 |             return _Up(unexpect, std::move(_M_unex));
      |                    ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/home/david/tests/cpp/src/expect.cpp: In function ‘int main()’:
/home/david/tests/cpp/src/expect.cpp:49:26: error: invalid use of ‘void’

What's the problem here? Is this feature only half done (so the standard is incomplete) or am I misunderstanding its point?

I modeled a function that returns an instance of an std::expected<void, Error> - I told myself "I can use a new standard so I will design my library accordingly" - and I was very positive to have nice and concrete error handling. Now it turns out, that all the monadic operations on the std::expected only work on non-void items. Even though there is a specialization for void, the monadic operations are not available. I understand that or_else needs to return the value - but there is a specialization for void, so why should this not work?

std::expected<void, Error> fun (int);

fun (19).and_then ([]() { doSomething(); }).or_else ([] (Error e) { std::println ("Uh oh error.."); });

This yields:

/usr/include/c++/14.2.1/expected:1586:37: error: static assertion failed
 1586 |           static_assert(__expected::__is_expected<_Up>);
      |                         ~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~
/usr/include/c++/14.2.1/expected:1586:37: note: ‘std::__expected::__is_expected<void>’ evaluates to false
/usr/include/c++/14.2.1/expected:1587:25: error: ‘std::remove_cvref<void>::type’ {aka ‘void’} is not a class, struct, or union type
 1587 |           static_assert(is_same_v<typename _Up::error_type, _Er>);
      |                         ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/usr/include/c++/14.2.1/expected:1592:20: error: expression list treated as compound expression in functional cast [-fpermissive]
 1592 |             return _Up(unexpect, std::move(_M_unex));
      |                    ^~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/home/david/tests/cpp/src/expect.cpp: In function ‘int main()’:
/home/david/tests/cpp/src/expect.cpp:49:26: error: invalid use of ‘void’

What's the problem here? Is this feature only half done (so the standard is incomplete) or am I misunderstanding its point?

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Share Improve this question Follow asked Nov 19, 2024 at 9:04 DavidDavid 39711 gold badge33 silver badges1212 bronze badges 3

• 3 The problem is that the callable to and_then and or_else needs to return a specialization of std::expected for chaining to work. Your lambdas do not. – StoryTeller - Unslander Monica Commented Nov 19, 2024 at 9:10
• Thanks for the hint - since the expected value is of type void - what should the callable to and_then return? – David Commented Nov 19, 2024 at 9:27
• I expanded my comment into an answer. – StoryTeller - Unslander Monica Commented Nov 19, 2024 at 9:34

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2 Answers 2

Sorted by: Reset to default 17

The problem is that the callable to and_then and or_else needs to return a specialization of std::expected for chaining to work. Your lambdas do not.

If fun were to return an unexpected (an error), and_then won't call your callable, but it will need to forward the error into the same sort of unexpected type that the lambda returns. So the lambda for and_then should return std::unexpected<..., Error>. You can choose what ever "expected" type you want the lambda to return, just as long as it's the same Error type. You can stick to void of course.

fun (19).and_then ([]() { doSomething(); return std::expected<void, Error>(); })

Conversly, the callbale to or_else can choose a different error type, but needs to preserve the expected type in its returned std::expected specialization. But all in all, the same treatment should work.

fun (19).and_then ([]() { 
          doSomething();
          return std::expected<void, Error>();
}).or_else ([] (Error e) { 
          std::println ("Uh oh error..");
          return std::expected<void, Error>();
});

Note I opted to have or_else swallow the error and return an std::expected with a value. You could choose to forward the error, of course.

...
.or_else ([] (Error e) { 
          std::println ("Uh oh error..");
          return std::expected<void, Error>(std::unexpect, e);
});

Although this is already answered by another answer to this question, let me add that transform and transform_error is probably what you expected for the library function instead.

fun(19).transform([]() { doSomething(); }).transform_error([](Error e) {
    std::println("Uh oh error..");
    return e;
});

There is a difference between monad and functor.

• and_then and or_else behaves monadically. and_then take in a function T -> std::expected<U, E> and transform from std::expected<T, E> to std::expected<U, E>
• transform and transform_error behave like functor. transform take in a function T -> U and transform from std::expected<T, E> to std::expected<U, E>.

It is interesting that the transform_error cannot have U = void if T is not void based on my testing. The monadic variant is more expressive as you can be returning a different result, for example returning the unexpected variant whenever something goes wrong inside and_then instead.

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