haskell - 为什么下面的代码无法解析？

Question

以下代码不解析:

main :: IO ()
main = do
    print $ result
        where result = foldl' (+) 0 [1..1000000]
    print $ result
        where result = last [1..1000000]

编译器在第二次打印时提示: src/Main.hs:10:5: 输入“print”时解析错误

这是为什么？

Answer 1

问题是 where 子句只能附加到绑定(bind)，不能附加到表达式。事实上:

main :: IO ()
main = do
    print $ result
        where result = foldl' (+) 0 [1..1000000]

完全等同于:

main :: IO ()
main = do
    print $ result
    where result = foldl' (+) 0 [1..1000000]

即where 定义了 main 的本地定义，而不是 print $ result 行。由于 where 必须是绑定(bind)的最后一部分，显然以下 print 表达式会导致语法错误。

要在 do block 中使用 where，您必须在定义 let 绑定(bind)时使用它，例如这个(非常愚蠢的示例):

main = do
    let result = f
        where f = foldl' (+) 0 [1..1000000]
    print result

---

您可以在 grammar 中查看:

decl    →   gendecl
        |   (funlhs | pat) rhs

rhs     →   = exp [where decls]
        |   gdrhs [where decls] 

请注意，where decls 是 rhs 规则的一部分，该规则定义了声明的右侧。如果您检查 exp 的规则，您将找不到提到的 where:

exp     →   infixexp :: [context =>] type       (expression type signature)
        |   infixexp

infixexp    →   lexp qop infixexp               (infix operator application)
            |   - infixexp                      (prefix negation)
            |   lexp

lexp    →   \ apat1 … apatn -> exp              (lambda abstraction, n ≥ 1)
        |   let decls in exp                    (let expression)
        |   if exp [;] then exp [;] else exp    (conditional)
        |   case exp of { alts }                (case expression)
        |   do { stmts }                        (do expression)
        |   fexp
fexp    →   [fexp] aexp                         (function application)

aexp    →   qvar                                (variable)
        |   gcon                                (general constructor)
        |   literal
        |   ( exp )                             (parenthesized expression)
        |   ( exp1 , … , expk )                 (tuple, k ≥ 2)
        |   [ exp1 , … , expk ]                 (list, k ≥ 1)
        |   [ exp1 [, exp2] .. [exp3] ]         (arithmetic sequence)
        |   [ exp | qual1 , … , qualn ]         (list comprehension, n ≥ 1)
        |   ( infixexp qop )                    (left section)
        |   ( qop⟨-⟩ infixexp )                 (right section)
        |   qcon { fbind1 , … , fbindn }        (labeled construction, n ≥ 0)
        |   aexp⟨qcon⟩ { fbind1 , … , fbindn }  (labeled update, n  ≥  1)