[英]Why this syntax error? No given reason from the shell
我剛剛寫完這段模式匹配(下面附上)。 無論我如何嘗試修復它,當我嘗試編譯它時仍然存在“錯誤:語法錯誤”。
這段代碼在這里:
let rec stmt (s:Ast.stmt) : X86.inst list =
let get_label e =
let inst = stmt e in
let pattern = (List.nth (List.rev inst) 0) in
let label = match pattern with
|Memory (Store (Addr (Label l), Reg r)) -> l
|_ -> "Error"
in
match s with
|Exp e -> (match e with
|Int i -> [movei RAX i; store new_temp() RAX]
|Var v -> [load RBX v; store new_temp() RBX]
|Binop (e1, op, e2) -> let (label1, label2) = (get_label e1, get_label e2)
in let operation =
match op with
|Plus -> [Arith (Add (RAX, Reg RBX))]
|Minus -> [Arith (Sub (RAX, Reg RBX))]
|Times -> [Arith (Mul RBX)]
|Div -> [Arith (Div RBX)]
|Eq -> [Arith (Cmp (RAX, RBX))]
|Neq -> [Arith (Cmp (RAX, Reg RBX)); set NE RAX]
|Lt -> [Arith (Cmp (RAX, Reg RBX)); set L RAX]
|Lte -> [Arith (Cmp (RAX, Reg RBX)); set LE RAX]
|Gt -> [Arith (Cmp (RAX, Reg RBX)); set G RAX]
|Gte -> [Arith (Cmp (RAX, Reg RBX)); set GE RAX]
in [load RAX label1; load RBX label2] @ operation @ [store (new_temp()) RAX]
|Not e -> let label = get_label e in
[load RBX label; Arith (Neg RBX); store (new_temp()) RAX]
|And (e1, e2) -> let (label1, label2) = (get_label e1, get_label e2) in
[load RAX label1; load RBX label2; Bitop (And (RAX, Reg RBX)); store (new_temp()) RAX]
|Or (e1, e2) -> let (label1, label2) = (get_label e1, get_label e2) in
[load RAX label1; load RBX label2; Bitop (Or (RAX, Reg RBX)); store (new_temp()) RAX]
|Assign (e1, e2) -> let label = get_label e2 in
[movei RAX label; store e1 RAX; store (new_temp()) RAX]
|_ -> [])
|Block e -> []
|If (l, m, r) -> []
|While (l, r) -> []
|For (l, m1, m2, r) -> []
|Return e -> []
|_ -> []
let compile (p : Ast.program) : result =
let _ = reset() in
let _ = collect_vars(p) in
let insts = List.concat_map stmt p in
{ code = insts; data = VarSet.elements !variables }
錯誤在於let compile (p: Ast.program): result =
這是(來自外殼):
110 | let compile (p : Ast.program) : result =
^^^
Error: Syntax error
錯誤源於未正確完成get_label
function:
let get_label e =
let inst = stmt e in
let pattern = (List.nth (List.rev inst) 0) in
let label = match pattern with
|Memory (Store (Addr (Label l), Reg r)) -> l
|_ -> "Error"
因此in
盡管您有縮進,但以下它與let label =...
,而不是let get_label e =...
相關聯。
我懷疑你想要的是跳過label
綁定:
let get_label e =
let inst = stmt e in
let pattern = (List.nth (List.rev inst) 0) in
match pattern with
| Memory (Store (Addr (Label l), Reg r)) -> l
| _ -> "Error"
使用像ocp-indent
這樣的工具可能是個好主意,它會根據編譯器解釋代碼的方式來縮進你的代碼,而不是你希望編譯器如何解釋它。 這通常有助於避免這樣的錯誤:)
聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.