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return
asyncSeq
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 }
a
AsyncSeq<int>
[1; 2]
3
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)
return! a
always do yield and yield!
yield
yield!
let a = asyncSeq { yield 1 yield 2 yield 3 } let b = asyncSeq { yield 4 yield! b }
The text was updated successfully, but these errors were encountered:
Related: #38
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@eulerfx I missed that. Closing this because is a duplicate.
Continuing discussion in other thread
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The
return
of theasyncSeq
computation ignore the value passed.I found it bug prone when upgrading a library from asyncseq v1 to v2 types.
a
isAsyncSeq<int>
[1; 2]
3
is ignored, no warning, no errorsSame when trying to return another
asyncSeq
likei think that should be
return! a
to work correctly (not yet supported)workaround
always do
yield
andyield!
The text was updated successfully, but these errors were encountered: