最外部的评估如何在应用咖喱函数时起作用？ [英] How does outermost evaluation work on an application of a curried function?

问题描述

mult 被定义为咖喱函数：

mult    ::  Int ->  Int ->  Int
mult    x   =   \y  ->  x   *   y

在 mult（1 + 2）（2 + 3） ，

• 什么是redex。它们是 mult（1 + 2）， 1 + 2 和 2 + 3 ？


• 什么是最外面的redex，它是 2 + 3 吗？

• what are the redex's. and are they mult(1+2), 1+2 and 2+3?
• What is the outermost redex, and is it 2+3?

最里面的评估按以下表达式工作：

Innermost evaluation works on the expression as following, according to Programming in Haskell by Hutton:

mult    (1+2)   (2+3)
= { applying    the first   +   }
mult    3   (2+3)
= { applying    mult    }
(\y ->  3   *   y)  (2+3)
= { applying    +   }
(\y ->  3   *   y)  5
= { applying    the lambda  }
3   *   5
= { applying    *   }
15

mult（1 + 2）（2 + 3）的最外部评估工作？
最外面的评估是否按以下方式工作？

How does outermost evaluation work on mult (1+2) (2+3)? Does outermost evaluation works as the following?

mult (1+2) (2+3)
= mult (1+2) 5
= (\y -> (1+2) * y) 5
= (1+2) * 5  // Is (1+2) evaluated before (1+2) * 5, because builtin function "*" is strict, i.e. application of builtin function always happen after evaluation of its args?
= 3*5
= 15

谢谢。

推荐答案

写下分析树：

         o
       /   \
     o       o
    / \     /|\
mult   o   2 + 3
      /|\
     1 + 2

（为简单起见，我正在处理二进制中缀 + 运算符作为单个应用程序，它也可能是（（+）1）2 ）

(For simplicitly I'm treating the binary infix + operators as a single application, it could have been ((+) 1) 2 as well)

现在最外层的函数应用是 mult（1 + 2）到参数 2 + 3 的应用，但它不可约，因为函数不是单个值，而是应用程序本身。我们必须首先进行评估：

Now the outermost function application is that of mult (1+2) to the argument 2+3, but it's not reducible because the function is not a single value but an application itself. We have to evaluate that first:

(mult (1+2)) (2+3)
((\x->\y->x*y) (1+2)) (2+3) -- the value that `mult` refers to
(\y->(1+2)*y) (2+3) -- evaluate the application of `\x->`

$ b $的应用b

现在我们可以评估根函数应用程序：

Now we can evaluate the root function application:

(1+2) * (2+3) -- application of `\y->`

现在最外面的表达式是 * ，但是您知道这些整数运算符很严格，因此需要首先评估其参数（从左到右，IIRC）：

Now the outermost expression is the *, but as you know these integer operators are strict so they need to evaluate their arguments first (left-to-right, IIRC):

3 * (2+3)
3 * 5
15

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