Javascript ..在本教程中完全丢失了 [英] Javascript..totally lost in this tutorial

问题描述

我的入门教程突然变得非常先进。我不知道这个程序是如何工作的。你能用简单的语言解释一下吗？

My introductory tutorial has suddenly become very advanced. I have no idea how this program works. Can you explain in plain language?

最后打印（（（1 * 3）+ 5）* 3），但我根本没有得到它。据我所知，findSequence传递了24，触发函数find。我假设函数find被传递1，1，后者被分配给历史？但我不明白为什么第二个1用引号1！，当我返回find时也不理解引号的使用（开始等等）

At the end it prints (((1 * 3) + 5) * 3), but I don't get it at all. I understand that findSequence gets passed 24, that triggers function find. I'm assuming that function find gets passed 1,"1" with the latter being assigned to history?? but I don`t understand why the second 1 is in quotation marks "1!, nor do I understand the use of quotation marks when it returns find(start etc.

function findSequence(goal) {
  function find(start, history) {
    if (start == goal)
      return history;
    else if (start > goal)
      return null;
    else
      return find(start + 5, "(" + history + " + 5)") ||
             find(start * 3, "(" + history + " * 3)");
  }
  return find(1, "1");
}

print(findSequence(24));

推荐答案

你有一个输出string （（（1 * 3）+ 5）* 3），目标 24 和第一个字符 1 。 1 是 1 * 3 中的1>这个程序将汇编一个包含数学表达式的字符串，添加（），* 3和+5以试图获得目标。 find 函数的两个参数是目前tal和将产生总数的表达式。显然，表达式是一个字符串。查找将当前总数与目标进行比较，如果它相等则表达式是正确的并返回它，如果当前总数>目标，则失败并返回null。另外，他在表达式中添加了一些操作。他尝试了两种方式，一种是将当前结果乘以* 3，另一种是将+5加到当前结果中。它在树上是递归的（因此每次他将在两个递归调用中分叉）。 null ||某事==某事，所以发现响应的分支将返回他的响应，另一分支将返回null并且获胜响应将被传回。

You have an "output" string (((1 * 3) + 5) * 3), a goal 24 and a first character 1. The 1 is the 1 in 1 * 3. This program will assemble a string containing a math expression, adding (), *3 and +5 to try to obtain the goal. The two parameters of the find function are the current total and the expression that will generate the total. Clearly the expression is a string. The find compares the current total to the goal, and if it's equal then the expression is correct and it returns it, if the current total is > of the goal then he failed and return null. Otherwhise he add some operations to the expression. He tries two "ways", one multiplying * 3 the current result, the other adding +5 to the current result. It's recursive on a tree (so each time he will bifurcate in two recursive calls). null || something == something, so the branch that will find a response will return his response, the other branch will return null and the "winning" response will be passed back.

假设目标是11。

• find（1，1）
  

• find(1, "1")

1. 将1与11进行比较并调用：（2。）find（1 + 5，（+1++ 5））（因此找到（6，（1 + 5））和（3。 ）find（1 * 3，（+1+* 3））（所以找（3，（1 * 3））


2. 将6与11进行比较并调用（4.）find（6 + 5，（+（1 + 5）++ 5））（因此找到（11，（（1 + 5）+ 5））和（ 5.）找到（6 * 3，（+（1 + 5）+* 3）（所以找到（18，（（1 + 5）* 3）


3. 将3与11进行比较并调用（6.）find（3 + 5，（+（1 * 3）++ 5））（所以找到（8，（（1 * 3） ）+ 5））和（7.）找到（3 * 3，（+（1 * 3）+* 3）（所以找（8，（（1 + 3）* 3）


4. 将11与11进行比较。数字相等。所以他返回（（1 + 5）+ 5）（历史）.5,6,7将确定点落水并超过11，所以th ey将返回null。 null || null == null，（（1 + 5）+ 5）|| null ==（（1 + 5）+ 5），因此历史记录将赢得空值并将返回。

1. compares 1 with 11 and calls: (2.) find(1 + 5, "(" + "1" + " + 5)") (so find(6, "(1 + 5)") and (3.) find(1 * 3, "(" + "1" + " * 3)") (so find(3, "(1 * 3)")
2. compares 6 with 11 and calls (4.) find (6 + 5, "(" + "(1 + 5)" + " + 5)") (so finds(11, "((1 + 5) + 5)") and (5.) find (6 * 3, "(" + "(1 + 5)" + " * 3)" (so find(18, "((1 + 5) * 3)"
3. compares 3 with 11 and calls (6.) find (3 + 5, "(" + "(1 * 3)" + " + 5)") (so finds(8, "((1 * 3) + 5)") and (7.) find (3 * 3, "(" + "(1 * 3)" + " * 3)" (so find(8, "((1 + 3) * 3)"
4. compares 11 with 11. The numbers are equal. So he returns "((1 + 5) + 5)" (the history). 5, 6, 7 will at a certain point go "overboard" and surpass the 11, so they'll return null. null || null == null, "((1 + 5) + 5)" || null == "((1 + 5) + 5)", so the history will win against the nulls and it will be returned.

为了使它更清晰，请尝试以下版本：

To make it even clearer, try these versions:

function findSequence(goal) {
  function find(start, history) {
    if (start == goal)
      return history;
    else if (start > goal)
      return null;
    else {
      var ret = find(start + 5, "(" + history + " + 5)");

      if (ret == null)
         ret = find(start * 3, "(" + history + " * 3)");

      return ret;
    }
  }

  return find(1, "1");
}

print(findSequence(24));

而且，除了表达式之外，你只能获得+和*

And this, where instead of an expression you'll get only as tring of + and *

function findSequence(goal) {
  function find(start, history) {
    if (start == goal)
      return history;
    else if (start > goal)
      return null;
    else {
      var ret = find(start + 5, history + "+");

      if (ret == null)
         ret = find(start * 3, history + "*");

      return ret;
     }
  }

  return find(1, "1");
}

print(findSequence(24));

请注意，作为一个例子，它非常复杂，因为它使用了闭包（本地定义的函数） 。

And be aware that, as an example, it's quite complex because it used closures (locally defined functions).

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