子函数隐式地使用递归，例如:

去迪士尼乐园自驾游

我们到了吗?(没有) 我们到了吗?(很快) 我们到了吗?(快了……) 我们到了吗? 我们到了吗?(!!!!!)

这时孩子就睡着了……

这个倒数函数是一个简单的例子:

倒计时()函数 ｛ 返回(参数[0]> 0 ? （ Console.log(参数[0])，倒计时(参数[0]- 1)): “完成” ）; ｝ 倒计时(10);

霍夫施塔特定律也适用于软件项目。

The essence of human language is, according to Chomsky, the ability of finite brains to produce what he considers to be infinite grammars. By this he means not only that there is no upper limit on what we can say, but that there is no upper limit on the number of sentences our language has, there's no upper limit on the size of any particular sentence. Chomsky has claimed that the fundamental tool that underlies all of this creativity of human language is recursion: the ability for one phrase to reoccur inside another phrase of the same type. If I say "John's brother's house", I have a noun, "house", which occurs in a noun phrase, "brother's house", and that noun phrase occurs in another noun phrase, "John's brother's house". This makes a lot of sense, and it's an interesting property of human language.

参考文献

递归与人类思想

递归

方法A调用方法A调用方法A，最终这些方法A中的一个不会调用并退出，但这是递归，因为有东西调用了它自己。

递归的例子，我想打印出硬盘驱动器上的每个文件夹名称:(在c#中)

public void PrintFolderNames(DirectoryInfo directory)
{
    Console.WriteLine(directory.Name);

    DirectoryInfo[] children = directory.GetDirectories();

    foreach(var child in children)
    {
        PrintFolderNames(child); // See we call ourself here...
    }
}

Ouch. I tried to figure out the Towers of Hanoi last year. The tricky thing about TOH is it's not a simple example of recursion - you have nested recursions which also change the roles of towers on each call. The only way I could get it to make sense was to literally visualize the movement of the rings in my mind's eye, and verbalize what the recursive call would be. I would start with a single ring, then two, then three. I actually ordered the game on the internet. It took me maybe two or three days of cracking my brains to get it.

要向一个六岁的孩子解释递归，首先要向一个五岁的孩子解释，然后再等一年。

实际上，这是一个有用的反例，因为递归调用应该更简单，而不是更难。向一个5岁的孩子解释递归就更难了，尽管你可以在0点停止递归，但你没有简单的解决方案来向一个0岁的孩子解释递归。

要使用递归解决一个问题，首先将其细分为一个或多个可以用相同方法解决的更简单的问题，然后当问题简单到无需进一步递归就可以解决时，您可以返回到更高的级别。

事实上，这是用递归方法来解决问题的递归定义。

Common Lisp中的简单递归示例:

MYMAP对列表中的每个元素应用一个函数。

1)空列表没有元素，所以我们返回空列表-()和NIL都是空列表。

2)将函数应用到第一个列表，对列表的其余部分调用MYMAP(递归调用)，并将两个结果合并到一个新列表中。

(DEFUN MYMAP (FUNCTION LIST)
  (IF (NULL LIST)
      ()
      (CONS (FUNCALL FUNCTION (FIRST LIST))
            (MYMAP FUNCTION (REST LIST)))))

让我们观察跟踪执行。在输入函数时，输出参数。在退出函数时，输出结果。对于每个递归调用，输出将按级别缩进。

这个例子对列表(1 2 3 4)中的每个数字调用SIN函数。

Command: (mymap 'sin '(1 2 3 4))

1 Enter MYMAP SIN (1 2 3 4)
| 2 Enter MYMAP SIN (2 3 4)
|   3 Enter MYMAP SIN (3 4)
|   | 4 Enter MYMAP SIN (4)
|   |   5 Enter MYMAP SIN NIL
|   |   5 Exit MYMAP NIL
|   | 4 Exit MYMAP (-0.75680256)
|   3 Exit MYMAP (0.14112002 -0.75680256)
| 2 Exit MYMAP (0.9092975 0.14112002 -0.75680256)
1 Exit MYMAP (0.841471 0.9092975 0.14112002 -0.75680256)

这是我们的结果:

(0.841471 0.9092975 0.14112002 -0.75680256)

如果你想要一本很好地用简单的术语解释递归的书，可以看看Gödel，埃舍尔·巴赫:道格拉斯·霍夫施塔特的《永恒的金辫子》，特别是第五章。除了递归，它还能很好地以一种可理解的方式解释计算机科学和数学中的许多复杂概念，一个解释建立在另一个解释的基础上。如果你以前没有接触过这类概念，这可能是一本非常令人兴奋的书。