回答編集履歴
3
eval の型について追記
test
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@@ -51,6 +51,26 @@
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その次の返り値 `assoc x list`の型は`'a option`と何かしらのオプション型なので
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型が全然合ってません。
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問題文を見ると、
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> eval [("x",4);("y",3)] (Mul(Mul(Var "x", Var "y"),Add(Var "x",Num 3)));;
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>
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> : int = Some 84
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なのですから `eval`の返り値は`int option`とすべきですね。
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つまり`|Num n -> n`の時点でもう型が違います。
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コード追加
test
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@@ -57,3 +57,115 @@
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とりあえず(全部消してイチから)全部の関数に型を明示して書いてはどうでしょう。
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---
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文法が似ているF#で書いてみました
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```F#
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open System
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type Exp =
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| Num of int
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| Var of string
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| Add of Exp * Exp
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| Mul of Exp * Exp
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// let hoge = Mul(Num 3, Var "x") // OK
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type VarValuePair = string * int
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let rec assoc (name : string) (pairs : VarValuePair list) : int option =
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match pairs with
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| [] -> None
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| (var, value)::t ->
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if var = name then Some value
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else assoc name t
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let rec eval (pairs : VarValuePair list) (exp : Exp): int option =
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match exp with
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| Num x -> Some x
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| Var var -> assoc var pairs
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| Add (x, y) ->
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let x = (eval pairs x)
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let y = (eval pairs y)
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match (x, y) with
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| (Some x, Some y) -> Some (x+y)
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| _ -> None
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| Mul (x, y) ->
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let x = (eval pairs x)
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let y = (eval pairs y)
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match (x, y) with
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| (Some x, Some y) -> Some (x*y)
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| _ -> None
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[<EntryPoint>]
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let main argv =
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let result = eval [("x",4); ("y",3)] (Mul(Mul(Var "x", Var "y"),Add(Var "x",Num 3)))
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printfn "%A" result
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0 // return an integer exit code
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```
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1
追記
test
CHANGED
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@@ -1 +1,59 @@
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1
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(x list)とカッコがついているのでxが関数と推論されているのでは？
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---
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`assoc x list`にすると型が合わないと怒られる問題について:
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```ocaml
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type exp =
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| Num of int
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| Var of string
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| Add of exp * exp
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| Mul of exp * exp;;
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//...
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let rec eval list var =
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match var with
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|Num n -> n
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|Var x -> assoc x list
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|Add (x, y) -> calc calc (eval (var));;
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```
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とあるので evalの最初のパターンマッチの返り値`n`は`int`ですが、
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|
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51
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+
その次の返り値 `assoc x list`の型は`'a option`と何かしらのオプション型なので
|
|
52
|
+
|
|
53
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+
型が全然合ってません。
|
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54
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+
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55
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+
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56
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+
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57
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+
|
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58
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+
|
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59
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+
とりあえず(全部消してイチから)全部の関数に型を明示して書いてはどうでしょう。
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