the return of asyncSeq builder is non intuitive

[enricosada]

The return of the asyncSeq computation ignore the value passed.
I found it bug prone when upgrading a library from asyncseq v1 to v2 types.

let a = asyncSeq {
    yield 1
    yield 2
    return 3
}

• no errors
• the a is AsyncSeq<int>
• as list, it's [1; 2]
• the 3 is ignored, no warning, no errors

Same when trying to return another asyncSeq like

let b = asyncSeq {
   yield 4
   return a
}

i think that should be return! a to work correctly (not yet supported)

workaround

always do yield and yield!

let a = asyncSeq {
    yield 1
    yield 2
    yield 3
}

let b = asyncSeq {
   yield 4
   yield! b
}

[eulerfx]

Related: #38

[enricosada]

@eulerfx I missed that. Closing this because is a duplicate.

Continuing discussion in other thread