C语言之递归

递归例子如下:

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

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

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

程序结果如下:

[zsd@TOMCAT ~]$ ./test03
level1-value of
level1-value of
level1-value of
level1-value of
level1-value of
level1-value of
level1-value of
level1-value of
level1-value of
---------------------------邪恶的分割线------------------------
level2-value of
level2-value of
level2-value of
level2-value of
level2-value of
level2-value of
level2-value of
level2-value of
level2-value of 

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 

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

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

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

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

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

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

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

Breakpoint , digui (n=) at test03.c:
          digui(n-);                 //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 

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

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

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

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

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

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

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

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

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

Program exited normally.