mguimas
What does %Mod{} really do?
Hello,
given the files xx.ex and yy.ex defined as
defmodule XX do
defstruct [:field]
end
defmodule YY do
defdelegate __struct__, to: XX
defdelegate __struct__(x), to: XX
end
why does
iex(1)> %XX{} == XX.__struct__()
true
and
iex(2)> %YY{} == YY.__struct__()
false
give different results?
It seems that %Mod{} is not equivalent to Mod.__struct__() …
What does %Mod{} really do?
Thanks
Mário
Most Liked
peerreynders
iex(14)> Map.to_list(%YY{})
[__struct__: YY, field: nil]
iex(15)> Map.to_list(YY.__struct__())
[__struct__: XX, field: nil]
Based on the evidence I suspect that the compiled code (it’s a special form after all) sanitizes the value returned from the function by forcing the __struct__ value to reflect the primary module - but that is just a guess.
peerreynders
iex(16)> struct!(YY,field: 1)
%XX{field: 1}
iex(17)> struct!(%YY{},field: 1)
%YY{field: 1}
I suspect the bug is with Kernel.struct!/2, not Kernel.SpecialForms.%/2.
As you discovered, it only presents when you start defdelegating the constructor and I don’t think this code anticipated that.
So the question is - what are you trying to accomplish by defdelegating the struct constructor to another struct? Sharing code between XX and YY should probably be accomplished via a shared third module.
Popular in Questions
Other popular topics
Latest on Elixir Forum
Sponsor Spotlight
Producing high quality Elixir screencasts since 2017.
Categories:
Sub Categories:
Forums
Our Sponsors
Build Elixir applications with speed and confidence.
Supporting innovation across the BEAM ecosystem.
We build reliable cloud platforms for business-critical systems.
Catch errors, track performance, monitor hosts and more.
Practical resources that improve the lives of professional developers.
Develop your skills with books, videos, and courses.
Enabling companies to succeed by building software people love.
Courses that'll move you from confusion to "Aha, now I get it!"