递归例子如下:
1 #include <stdio.h> 2 /*函数声明*/ 3 void digui(int n); 4 5 int main() 6 { 7 int n=10; 8 digui(n); 9 return 0; 10 } 11 12 void digui(int n) 13 { 14 printf("level1-value of %d\n",n); 15 if(n>2){ 16 digui(n-1); 17 } 18 printf("level2-value of %d\n",n); 19 }
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.
