C语言之递归

作者：@张扶摇
本文为作者原创，转载请注明出处：https://www.cnblogs.com/zhangshengdong/p/9857034.html

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目录

递归例子如下:

#include <stdio.h>
/*函数声明*/
void digui(int n);

int main()
{
  int n=10;
  digui(n);
  return 0;
}

void digui(int n)
{
  printf("level1-value of %d\n",n);
  if(n>2){
    digui(n-1);
  } 
  printf("level2-value of %d\n",n);
}

 

程序结果如下:

[zsd@TOMCAT ~]$ ./test03
level1-value of 10
level1-value of 9
level1-value of 8
level1-value of 7
level1-value of 6
level1-value of 5
level1-value of 4
level1-value of 3
level1-value of 2
---------------------------邪恶的分割线------------------------
level2-value of 2
level2-value of 3
level2-value of 4
level2-value of 5
level2-value of 6
level2-value of 7
level2-value of 8
level2-value of 9
level2-value of 10

通过gdb的调试，对代码的16行和18行设置断点，gdb执行的效果如下:

(gdb) run
Starting program: /home/zsd/test03debug 
level1-value of 10

Breakpoint 1, digui (n=10) at test03.c:16
16          digui(n-1);                  //开始第一次向下递归，递归数为9
(gdb) 
(gdb) continue
Continuing.
level1-value of 9

Breakpoint 1, digui (n=9) at test03.c:16
16          digui(n-1);                 //向下递归，递归数为8
(gdb) continue
Continuing.
level1-value of 8

Breakpoint 1, digui (n=8) at test03.c:16
16          digui(n-1);
(gdb) continue
Continuing.
level1-value of 7

Breakpoint 1, digui (n=7) at test03.c:16
16          digui(n-1);
(gdb) continue
Continuing.
level1-value of 6

Breakpoint 1, digui (n=6) at test03.c:16
16          digui(n-1);
(gdb) continue
Continuing.
level1-value of 5

Breakpoint 1, digui (n=5) at test03.c:16
16          digui(n-1);
(gdb) continue
Continuing.
level1-value of 4

Breakpoint 1, digui (n=4) at test03.c:16
16          digui(n-1);
(gdb) continue
Continuing.
level1-value of 3

Breakpoint 1, digui (n=3) at test03.c:16
16          digui(n-1);                 //一直到这里，递归数为2.这个时候，上面递归的每一个函数digui(2),digui(3)....digui(10)有最后一条printf("level2-value of %d\n",n);语句没有执行。
(gdb) continue
Continuing.
level1-value of 2

Breakpoint 2, digui (n=2) at test03.c:18   //digui(2)执行printf("level2-value of %d\n",n);语句
18        printf("level2-value of %d\n",n);
(gdb) continue
Continuing.
level2-value of 2

Breakpoint 2, digui (n=3) at test03.c:18   //digui(3)执行printf("level2-value of %d\n",n);语句
18        printf("level2-value of %d\n",n);
(gdb) continue
Continuing.
level2-value of 3

Breakpoint 2, digui (n=4) at test03.c:18   //以上述递推，一直到digui(10)执行完毕。
18        printf("level2-value of %d\n",n);
(gdb) continue
Continuing.
level2-value of 4

Breakpoint 2, digui (n=5) at test03.c:18
18        printf("level2-value of %d\n",n);
(gdb) continue
Continuing.
level2-value of 5

Breakpoint 2, digui (n=6) at test03.c:18
18        printf("level2-value of %d\n",n);
(gdb) continue
Continuing.
level2-value of 6

Breakpoint 2, digui (n=7) at test03.c:18
18        printf("level2-value of %d\n",n);
(gdb) continue
Continuing.
level2-value of 7

Breakpoint 2, digui (n=8) at test03.c:18
18        printf("level2-value of %d\n",n);
(gdb) continue
Continuing.
level2-value of 8

Breakpoint 2, digui (n=9) at test03.c:18
18        printf("level2-value of %d\n",n);
(gdb) continue
Continuing.
level2-value of 9

Breakpoint 2, digui (n=10) at test03.c:18
18        printf("level2-value of %d\n",n);
(gdb) continue
Continuing.
level2-value of 10

Program exited normally.

 

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