对于每个到 XSL - 变量?递归?

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问题描述

我不习惯函数式编程，并且很难为每个循环找出一个简单的方法.

I am not used to functional programming and having hard times figuring out a simple for each loops.

我手头有这个，我需要在 xsl 中修正相同的算法.

I have this in hand and i need to right the same algorithm in xsl.

let the input be a string S consisting of n characters: a1 ... an.
let the grammar contain r nonterminal symbols R1 ... Rr.
This grammar contains the subset Rs which is the set of start symbols.
let P[n,n,r] be an array of booleans. Initialize all elements of P to false.
for each i = 1 to n
  for each unit production Rj -> ai
    set P[i,1,j] = true
for each i = 2 to n -- Length of span
  for each j = 1 to n-i+1 -- Start of span
    for each k = 1 to i-1 -- Partition of span
      for each production RA -> RB RC
        if P[j,k,B] and P[j+k,i-k,C] then set P[j,i,A] = true
if any of P[1,n,x] is true (x is iterated over the set s, where s are all the indices for Rs) then
  S is member of language
else
  S is not member of language

我可以做一些事情，但我觉得卡住了，走错了路.

I am able to do some but i feel stuck and going into the wrong way.

我了解到我不能拥有数组之类的东西.即使我创建了一个像数组一样的变量，我也无法更新它.

I learned that I cannot have something like arrays. Even i create a variable which acts like an array i cannot update it.

所以我开始做一些递归(通过调用模板)，但我仍然在考虑这是否是正确的方法.如果还有另一个原因，为什么要在没有递归的情况下为 eachs 实现嵌套..

So i started doing some recursions (by calling templates) but i am still thinking if it is the right way. And if there is another why to implement those nested for eachs without recursion..

FXSL

对于每个到 XSL - 变量?递归?

问题描述

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