[英]Curry in Haskell
as you know the foldl function is defined as : 如您所知,foldl函数定义为:
foldl :: (a -> b -> a ) -> a -> [b] -> a
I want to rewrite the function as an uncurrified function 我想将该函数重写为未使用的函数
I tried this one: 我尝试了这个:
foldl :: ( (a-> b-> a) , a , [b] ) -> a
Is that correct? 那是对的吗? Maybe it is not important to uncurry but I am gonna write an exam and I am pretty sure this will be one of the tasks to do. 也许通俗易懂并不重要,但我要写考试,而且我很确定这将是要做的任务之一。
Thanks in anticipation ! 谢谢您的期待!
Well, that sure is an uncurried form of foldl
. 好吧,那肯定是未经处理的foldl
形式。 However, there a more levels on which you can do this – what I'd can the "fully uncurried form" would be 但是,您可以在更高的级别上执行此操作–“完全未经处理的表单”应该是
foldl'' :: ( ((a,b) -> a), a, [b] ) -> a
where not just the function itself but also its function argument is uncurried. 不仅函数本身,而且函数参数都不是未处理的。 OTOH, just calling uncurry
on the function would yield merely OTOH,仅在函数上调用uncurry
只会产生
foldl''' :: ( (a->b->a), a ) -> [b] -> a
which might thus also be called "uncurried foldl
", though it would certainly not be the desired interpretation in an exam. 这可能因此也被称为“uncurried foldl
”,但它肯定不会在考试所需的解释。
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