我是 OCaml 的新手。我写了这段代码来减少代数表达式:
type expr =
| Int of int
| Float of float
| Add of expr*expr
| Sub of expr*expr
| Mult of expr*expr
| Div of expr*expr
| Minus of expr
let rec eval expression = match expression with
| Add (e1, e2) -> (eval e1) +. (eval e2)
| Sub (e1,e2) -> (eval e1) -. (eval e2)
| Mult (e1,e2) -> (eval e1) *. (eval e2)
| Div (e1, e2) -> (eval e1) /. (eval e2)
| Minus (e1) -> -.(eval e1)
| Int i -> (float) i
| Float f -> f
let rec simplify_expr e = match e with
| Add (e1,e2) -> if (eval e1) == 0.0 then simplify_expr e2
else if (eval e2) == 0.0 then simplify_expr e1
else Add (simplify_expr e1, simplify_expr e2)
| Mult(e1,e2) -> if (eval e1) == 1.0 then simplify_expr e2
else if (eval e2) == 1.0 then simplify_expr e1
else Mult (simplify_expr e1, simplify_expr e2)
| Sub (e1, e2) -> if (eval e1) == 0.0 then simplify_expr e2
else if (eval e2) == 0.0 then simplify_expr e1
else Sub (simplify_expr e1, simplify_expr e2)
| Div (e1, e2) -> if (eval e1) == 1.0 then simplify_expr e2
else if (eval e2) == 1.0 then simplify_expr e1
else Div (simplify_expr e1, simplify_expr e2)
| Int i -> e
| Minus e1 -> simplify_expr(e1)
| Float f -> e
我这样调用 simplify_expr:
Expr.simplify_expr Expr.Mult (Expr.Int 4, Expr.Add (Expr.Int 1, Expr.Int 0));;
我的答案是错误的:
- : Expr.expr = Expr.Mult (Expr.Int 4, Expr.Add (Expr.Int 1, Expr.Int 0))
下面我粘贴调用的堆栈。
Expr.simplify_expr <--
Expr.Mult (Expr.Int 4, Expr.Add (Expr.Int 1, Expr.Int 0))
Expr.eval <-- Expr.Int 4
Expr.eval --> 4.
Expr.eval <-- Expr.Add (Expr.Int 1, Expr.Int 0)
Expr.eval <-- Expr.Int 0
Expr.eval --> 0.
Expr.eval <-- Expr.Int 1
Expr.eval --> 1.
Expr.eval --> 1.
Expr.simplify_expr <-- Expr.Add (Expr.Int 1, Expr.Int 0)
Expr.eval <-- Expr.Int 1
Expr.eval --> 1.
Expr.eval <-- Expr.Int 0
Expr.eval --> 0.
Expr.simplify_expr <-- Expr.Int 0
Expr.simplify_expr --> Expr.Int 0
Expr.simplify_expr <-- Expr.Int 1
Expr.simplify_expr --> Expr.Int 1
Expr.simplify_expr --> Expr.Add (Expr.Int 1, Expr.Int 0)
Expr.simplify_expr <-- Expr.Int 4
Expr.simplify_expr --> Expr.Int 4
Expr.simplify_expr -->
Expr.Mult (Expr.Int 4, Expr.Add (Expr.Int 1, Expr.Int 0))
- : Expr.expr = Expr.Mult (Expr.Int 4, Expr.Add (Expr.Int 1, Expr.Int 0))
我不知道,为什么在带有 1(第 9 行)的 eval 返回后,用 Add 调用了 simplify_expr。有人可以帮忙吗?
最佳答案
将 ==
替换为 =
。参见例如 Does != have meaning in OCaml? .
不是问题,还要检查 simplify_expr
的 Sub
、Div
和 Minus
情况:0 - e2
不是 e2
并且 1/e2
不是 e2
...
关于recursion - OCaml 递归函数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13100503/