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content/blog/polymorphic-functions-wildcard-matching-lean-4.md

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+---
+title: "Polymorphic Functions w/ Wildcard Matching in Lean 4"
+date: 2024-08-04T07:21:07-07:00
+draft: false
+tags: []
+math: false
+medium_enabled: false
+---
+
+I was reading through the [Polymorphism](https://lean-lang.org/functional_programming_in_lean/getting-to-know/polymorphism.html) section in the Functional programming in Lean textbook and came across the following example:
+
+```lean4
+inductive Sign where
+  | pos
+  | neg
+
+def posOrNegThree (s : Sign) : match s with | Sign.pos => Nat | Sign.neg => Int :=
+  match s with
+  | Sign.pos => (3 : Nat)
+  | Sign.neg => (-3 : Int)
+```
+
+The function `posOrNegThree` depending on the input, can either return an expression of type `Nat` or an expression of type `Int`. That's super neat!
+
+What happens if we add a wildcard to the match? 
+
+For example, let's say we want a type based on the number of bits of precision specified. If we don't support the number of bits, we return the arbitrary precision `Nat` as our default.
+
+```lean4
+def UIntN (n: Nat) : Type := match n with
+  | 32 => UInt32
+  | 64 => UInt64
+  | _ => Nat
+```
+
+Now let's write a function that returns the zero element of our specified type:
+
+```lean4
+def u0 (x: Nat) : UIntN x := match x with
+  | 32 => (0: UInt32)
+  | 64 => (0: UInt64)
+  | _ => Nat.zero
+```
+
+This will result in the following error:
+
+```lean4
+type mismatch
+  Nat.zero
+has type
+  ℕ : Type
+but is expected to have type
+  UIntN x✝ : Type
+```
+
+I got stuck on this for a while, so I asked on the really helpful Lean Zulip and got a [great response](https://leanprover.zulipchat.com/#narrow/stream/113489-new-members/topic/.E2.9C.94.20Polymorphic.20Functions.20w.2F.20Wildcard.20Match.20Type)
+
+```lean4
+def u0 (x: Nat) : UIntN x := dite (x=32) (λ h ↦ by
+    subst h
+    exact (0 : UInt32)
+  ) (dite (x = 64) (λ h ↦ by
+    intro h2
+    subst h
+    exact (0: UInt64)
+    )
+    (λ h ↦ by
+    intro h2
+    have : UIntN x = Nat := by
+      unfold UIntN
+      simp only
+    rw [this]
+    exact 0
+  ))
+```
+
+Unfortunately this doesn't look pretty, but this was a massive clue in finding the prettier syntax to solve this problem!
+
+```lean4
+def u0 (x: Nat) : UIntN x :=
+  if h: x = 6 then by
+    subst h
+    exact (0: UInt32)
+  else if h2: x = 4 then by
+    subst h2
+    exact (0: UInt64)
+  else by
+    have : UIntN x = Nat := by
+      unfold UIntN
+      simp only
+    rw [this]
+    exact 0
+```
+
+
+