Haskell-这是关闭吗？ [英] Haskell - is this a closure?

问题描述

有一段源代码源于对我的另一个问题，

infFromPrefix :: Eq a => ([a] -> [a]) -> [a] -> [a] 
infFromPrefix rules prefix = inf where
    inf = prefix ++ case stripPrefix prefix (rules inf) of
        Just suffix -> suffix
        Nothing     -> error "Substitution does not preserve prefix"

我很确定 inf 必须是一个闭包，因为它可以使用其传递给 infFromPrefix 的参数来访问其封闭范围内的变量，但是不确定本质上 infFromPrefix 和 inf 是相同的函数， inf 只允许
a更简洁的定义。等效的定义为

where I am pretty sure that inf must be a closure as it has access to variables from its enclosing scope in the sense that it uses the parameters passed to infFromPrefix, but am unsure since essentially infFromPrefix and inf is the same function, the inf only allows a more succinct definition. An equivalent definition would be

infFromPrefix rules prefix = prefix ++ case stripPrefix prefix (rules $ infFromPrefix rules prefix) of
        Just suffix -> suffix
        Nothing     -> error "Substitution does not preserve prefix"

我正确，是 inf 关闭？

推荐答案

基于Wiki文章编程中的关闭，我认为可以说 inf 是确实不是闭包：

Based on the Wiki article on Closures in programming, I think it can be said that inf is indeed not a closure:

请特别注意，嵌套函数定义本身不是闭包：它们具有自由变量，尚未绑定。只有在使用参数的值对封闭函数进行求值后，嵌套函数绑定的自由变量才会创建闭包，然后从封闭函数中返回闭包。

Note especially that the nested function definitions are not themselves closures: they have a free variable, which is not yet bound. Only once the enclosing function is evaluated with a value for the parameter is the free variable of the nested function bound, creating a closure, which is then returned from the enclosing function.

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