理解递归如何在 C 中工作的问题

问题描述

我正在从C 编程:一种现代方法一书中学习 C 中的函数.它包含以下功能:

I'm learning about functions in C from the book C Programming: A Modern Approach. It contains the following function:

int fact(int n)
{
  if (n <= 1)
    return 1;
  else
    return n * fact(n - 1);
}

以下解释:

要了解递归是如何工作的，让我们跟踪执行语句 i = fact(3); 这里是发生了什么:

fact(3) finds that 3 is not equal or less than 1, so it calls
   fact(2)which finds that 2 is not equal or less than 1, so it calls
      fact(1) which finds that 1 is equal or less than 1, so it returns 1,causing
   fact(2) to return  2 x 1 = 2, causing
fact(3) to return 3 x 2 = 6

我真的很困惑，如果它不在循环内，为什么它会执行所有这些操作?为什么它一直使 3 越来越少，直到它与 if 函数中的参数匹配?

I'm really confused, why does it does all these actions if it is not inside a loop? why does it keep making the 3 less and less until it matches the argument in if function?

* 符号是否表示事实 (n) 中的 n 现在是 3 - 1 并且它基本上是说它又要重新做一遍?

Does the * symbol means that n in the fact(n) is now 3 - 1 and it basically says it do it all over again?

推荐答案

递归是函数调用自身的过程.在函数定义中，您可以看到对 fact(n-1) 的调用，这意味着再次调用函数本身.

Recursion is the process when a function calls itself.In the function definition you can see a call to fact(n-1), this means the function itself is called again.

int fact(int n) //return value is an integer, name of function is "fact", input argument is an integer named n
{
  if (n <= 1) // if the input arg equals 1, return 1
    return 1;
  else 
    return n * fact(n - 1); //else, return n (input arg of this fact) times the return value of fact(n - 1)
}

Fact(3) 应该返回 3 * 2 * 1 = 6.处理语句 int a = fact(3) 时会发生什么:

Fact(3) is supposed to return 3 * 2 * 1 = 6. What happens when process the statement int a = fact(3) is the following:

fact(3) //dive into this function:
{
  if (n <= 1) //n is not 1
    return 1;
  else
    return n * fact(n - 1); //now we call fact(2)
}

我们称之为事实(2):

we call fact(2):

{
  if (n <= 1) //n is not 1
    return 1;
  else
    return n * fact(n - 1); //now we call fact(1)
}

我们称之为事实(1):

we call fact(1):

{
  if (n <= 1) //n equals 1
    return 1; //return 1
  else
    return n * fact(n - 1); 
}

现在函数 fact(1) 返回 1 到它被调用的点，它在 fact(2) 内部.准确地说，对于语句return 2*fact(1);，这将被传递回语句return 3*fact(2);，它被传递回我们的初始函数调用和我们的整数 'a' 得到值 6.

Now function fact(1) returns 1 to the point where its called, which is inside fact(2).To be precise, to the statement return 2*fact(1);, this gets passed back to the statement return 3*fact(2); which gets passed back to our initial function call and our integer 'a' gets the value 6.

我希望这对您来说更清楚，如果您仍有问题，请随时提出.

I hope this made it more clear to you, if you have still questions left feel free to ask.

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