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aetherspawn
3078 days ago
Haskell has had the dot (.) operator which does the same thing for probably longer, but I have no reference sorry. Haskell is >30 years old.
2 comments
yorwba
3078 days ago
Dot has the arguments swapped.
f |> g = g . f
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scns
3077 days ago
The $-operator is Haskell's equivalent to |>
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