理解遞歸

 1 #include <iostream>
 2 using namespace std;
 3 int abc(int);
 4 int fun1(int);
 5 int fun1(int a)
 6 {
 7     cout << "遞歸函數1" << endl;
 8     if (a == 1)
 9     {
10         return 1;
11     }    
12     return a * fun1(a - 1);
13 }
14 int abc(int a)
15 {
16     cout << "遞歸函數0" << endl;
17     if (a == 1)
18     {
19         return 1;
20     }
21 
22     return a + abc(a - 1);
23 }
24 
25 int main()
26 {
27     int b = abc(4);
28     cout << b << endl;
29     int d = fun1(3);
30     cout << "原函數" << endl;    
31     cout << d << endl;
32     return 0;
33 }

運行結果:

遞歸函數0
遞歸函數0
遞歸函數0
遞歸函數0
10
遞歸函數1
遞歸函數1
遞歸函數1
原函數
6