为什么即使我指定 is 作为类型限制，GHC 也会抱怨缺少 `Typeable` 实例？

Question

即使我添加了边界Typeable来满足cast函数的要求，为什么以下代码也无法编译？

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import           Data.Data\n\na :: Maybe Int\na = cast f\n  where\n    f :: Typeable a => a\n    f = undefined\n

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app/Main.hs:6:5: error:\n    \xe2\x80\xa2 No instance for (Typeable a0) arising from a use of \xe2\x80\x98cast\xe2\x80\x99\n    \xe2\x80\xa2 In the expression: cast f\n      In an equation for \xe2\x80\x98a\xe2\x80\x99:\n          a = cast f\n            where\n                f :: Typeable a => a\n                f = undefined\n  |\n6 | a = cast f\n  |     ^^^^\n

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Maybe Int应该实现类型类，因为以下代码可以编译。

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app/Main.hs:6:5: error:\n    \xe2\x80\xa2 No instance for (Typeable a0) arising from a use of \xe2\x80\x98cast\xe2\x80\x99\n    \xe2\x80\xa2 In the expression: cast f\n      In an equation for \xe2\x80\x98a\xe2\x80\x99:\n          a = cast f\n            where\n                f :: Typeable a => a\n                f = undefined\n  |\n6 | a = cast f\n  |     ^^^^\n

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（我不是问如何转换一个值，例如从Int到String。我只是对这个奇怪的错误感到好奇。）

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GHC：9.2.2

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Answer 1

f问题是您从未在强制转换之前指定应具有的类型。你同样可以这样做

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a :: Maybe Int\na = cast (f :: Int)\n where f :: Typeable a => a\n       f = undefined\n

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（这会产生Just \xe2\x8a\xa5）或

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a :: Maybe Int\na = cast (f :: [String])\n where f :: Typeable a => a\n       f = undefined\n

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（这会产生Nothing）。

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某些约束会导致编译器默认未指定的类型，例如Num上下文默认为Integer. 但Typeable没有默认任何内容，因此编译器会犹豫。

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