Typed Return Type for `flatmap` contradicts documentation

[mrwizard82d1]

The documentation for flatmap states: "This is different than Option.map() because the result of the callback isn’t wrapped in a new Option." This test verifies that behavior:

    def test_flatmap_some():
        assert_that(option.Some(4).flatmap(lambda x: x + 1), equal_to(5))

However, the return type of the function is specified to be Option[U].

What is the expected behavior flatmap?

[mrwizard82d1]

I observed a similar discrepancy for option.Result.flatmap.

[MaT1g3R]

Hi, the documentation might be confusing. The intended purpose for flatmap is to have the call back you provide return an Option/Result, but for the final value be flattened into a single Option/Result. This is useful for chaining multiple computations that can fail/produce NONE with fail fast semantics.

Suggestions for new wording on the doc is welcome.

For example,

option.Some(5).map(lambda x: option.Some(x + 1)) == Some(Some(6))

but

option.Some(5).flatmap(lambda x: option.Some(x + 1)) == Some(6)

[mrwizard82d1]

Thank you for the response. Unfortunately, I’m swamped dealing with family issues and cannot focus on this issue right now. Hopefully, I will remember your response when I’m “unswamped.” Have a great day!

…

On Mar 30, 2022, at 3:04 PM, MaT1g3R ***@***.***> wrote: Hi, the documentation might be confusing. The intended purpose for flatmap is to have the call back you provide return an Option/Result, but for the final value be flattened into a single Option/Result. This is useful for chaining multiple computation that can fail/produce NONE with fail fast semantics. Suggestions for new wording on the doc is welcome. For example, option.Some(5).map(lambda x: option.Some(x + 1)) == Some(Some(6)) but option.Some(5).flatmap(lambda x: option.Some(x + 1)) == Some(6) — Reply to this email directly, view it on GitHub <#6 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAERLXXHGAFE2FN5HENB6ILVCSXT3ANCNFSM43L2GVMQ>. You are receiving this because you authored the thread.

[MaT1g3R]

@mrwizard82d1 thanks, wish you the best on your family issues

[mrwizard82d1]

Thank you very much.

…

On Mar 30, 2022, at 3:11 PM, MaT1g3R ***@***.***> wrote: @mrwizard82d1 <https://github.com/mrwizard82d1> thanks, wish you the best on your family issues — Reply to this email directly, view it on GitHub <#6 (comment)>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AAERLXVPWLYUYA2QV7T74JLVCSYO7ANCNFSM43L2GVMQ>. You are receiving this because you were mentioned.