Question

以下代码将无法编译：

import qualified Data.Vector.Unboxed as V
import qualified Data.Vector.Unboxed.Mutable as MV

bad :: V.Vector Int
bad = id . V.create $ do
    v <- MV.new 1
    MV.set v 0
    pure v

• Couldn't match type ‘m0 (MV.MVector
                             (Control.Monad.Primitive.PrimState m0) a0)’
                 with ‘forall s. GHC.ST.ST s (MV.MVector s Int)’
  Expected type: m0 (MV.MVector
                       (Control.Monad.Primitive.PrimState m0) a0)
                 -> V.Vector Int
    Actual type: (forall s. GHC.ST.ST s (MV.MVector s Int))
                 -> V.Vector Int
• In the second argument of ‘(.)’, namely ‘V.create’
  In the expression: id . V.create
  In the expression:
    id . V.create
    $ do { v <- MV.new 1;
           MV.set v 0;
           pure v }

但接下来将是：

import qualified Data.Vector.Unboxed as V
import qualified Data.Vector.Unboxed.Mutable as MV


good :: V.Vector Int
good = id $ V.create $ do
    v <- MV.new 1
    MV.set v 0
    pure v

stack解析器是lts-9.12

那么，如果我喜欢按(.)撰写，为什么会这样做以及如何修复？

Answer 1

你已经发现GHC无法做出不可预测的多态性这一事实。 $objectManager = \Magento\Framework\App\ObjectManager::getInstance(); $cart = $objectManager->get('\Magento\Checkout\Model\Cart'); // get cart items $items = $cart->getItems(); $isyes="0"; // get product attribute value of cart items foreach ($items as $item){ if(!$item->getData('test')=="yes"){ $isyes="1"; } }; if($isyes=="1"){ unset($jsLayout['components']['checkout']['children']['steps']['children']['shipping-step']['children'] ['shippingAddress']['children']['shipping-address-fieldset']['children']['test']); }的排名类型V.create较高，当您尝试将其应用于forall a. Unbox a => (forall s. ST s (MVector s a)) -> Vector a时，您最终需要实例化(id .) :: (x -> y) -> (x -> y)，这会爆炸。并不是说有任何真正错误的所需的实例化;只是GHC无法处理它。曾经有一个扩展名x = forall s. ST s (MVector s a)试图提供有限的支持，但是从GHC 8开始，它已经被彻底破坏和弃用了。 ImpredicativePolymorphism仍然有效的原因是因为它实际上已经硬连线到GHC中，因此它可以避免这个问题，这就是为什么如果你试图定义

($)

并在f $$ x = f x处使用它而不是($)。

使用向量`create` function

1 个答案: