Haskell`seq`运营商的时间成本 [英] Time cost of Haskell `seq` operator

问题描述

此常见问题解答表示

seq运算符是

  seq :: a  - > b  - > b 
  

x seq 足以检查它是不是底部，然后
丢弃结果并评估y。这看起来似乎不太有用，但它
意味着x在保证y被考虑之前保证被评估。

这太糟糕了Haskell不错，但这是否意味着在

  x`seq` fx 
  

评估 x 的费用将支付两次（放弃结果）？ seq 函数将放弃<$ c $>的值。 c> x ，但是由于已经评估过该值，因此所有对 x 的引用都将更新为不​​再指向未评估版本的 x ，而是指向评估版本。因此，尽管 seq 评估并放弃了 x ，但其值为 x ，因此不会重复评估。

This FAQ says that

The seq operator is

seq :: a -> b -> b

x seq y will evaluate x, enough to check that it is not bottom, then discard the result and evaluate y. This might not seem useful, but it means that x is guaranteed to be evaluated before y is considered.

That's awfully nice of Haskell, but does it mean that in

x `seq` f x

the cost of evaluating x will be paid twice ("discard the result")?

解决方案

The seq function will discard the value of x, but since the value has been evaluated, all references to x are "updated" to no longer point to the unevaluated version of x, but to instead point to the evaluated version. So, even though seq evaluates and discards x, the value has been evaluated for other users of x as well, leading to no repeated evaluations.

这篇关于Haskell`seq`运营商的时间成本的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！