如何分析一个递归源$ C ​​$ CS手？

问题描述

我如何分析一个递归源$ C ​​$ CS手？

例如，我设计了一种技术，通过手工分析迭代源$ C ​​$ C是这样的：

  INT事实（INT N）
{
 INT F = 0;
 INT I = 0;
 如果（N下; = 1）
 {
  返回1;
 } F = 1;
 我= 2;
 对于（i = 2; I＆LT; = N;我++）
 {
  F * = I;
 } 返回F;
}---------------------------
  I F新-F
---------------------------
---------------------------
---------------------------
 

对于每一个我，我可以分析和手工计算旧-F和新-F的值并填写表格看，如果日常工作正常。

但我怎么可以用手分析递归程序？

  INT事实（INT编号）{
    INT温度;    如果（数字＆LT; = 1）返回1;    TEMP =号*事实（ -  1）;
    返回温度;
}
 

解决方案

由于堆栈上的递归值存储你需要分析它的2路

第1次是：执行递归直到达到终止条件

。

第2次：直到堆栈为空收集值

。

  1日下降了2次
---------------------------------
N = 6 TMP = 6 * 120 = 720＆LT;  - 结果
N = 5 TMP = 5 * 24 = 120
n = 4的TMP = 4 * 6 = 24
n = 3的TMP = 3 * 2 = 6
n = 2的TMP = 2 * 1 = 2
N = 1结束
 

How can I analyze a recursive source codes by hand?

For example, I have devised a technique for analyzing iterative source code by hand like this:

int fact(int n) 
{ 
 int f = 0; 
 int i = 0; 
 if (n<=1)  
 {  
  return 1; 
 } 

 f = 1; 
 i = 2; 
 for (i=2; i<=n ; i++) 
 { 
  f *= i; 
 } 

 return f; 
}

---------------------------
  i         f       new-f
---------------------------
---------------------------
---------------------------

For each 'i' I can analyze and calculate the values of old-f and new-f by hand and fill up the table to see if the routine is working correctly.

But how can I analyze recursive routines by hand?

int fact(int number) {
    int temp;

    if(number <= 1) return 1;

    temp = number * fact(number - 1);
    return temp;
}

解决方案

Since recursion stores values on the stack you need to analyze it 2-way

1st pass is: do the recursion until the termination condition is reached.

2nd pass: collect the values until the stack is empty.

1st down  2nd up
---------------------------------
n = 6     tmp = 6 * 120 = 720 <- result
n = 5     tmp = 5 * 24 = 120
n = 4     tmp = 4 * 6 = 24
n = 3     tmp = 3 * 2 = 6
n = 2     tmp = 2 * 1 = 2
n = 1     end

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