关于C语言海龟图问题

#include <stdio.h>
#include <stdlib.h>
char state = "a";
int i=0,j=0,xstate=0,Dstate[1][2]={{0,1}};
int floor[50][50]={0};
int direction( int m1);
int move(int n1,int x1);
int print(void);
int main()
{ 
  int x,m,n,wind=1;
  int i1=0,j1=0;
     for(i1=0;i1<25;i1++)
      {for(j1=0;j1<50;j1++)
        {  
           printf("%d",floor[i1][j1]=0);
           }  
           printf("\n");
        }//将floor这个数组的前25*50都置为0
  
  while (wind!=0)   /*当wind为0时退出*/ 
  {
  printf("\ninput the comand:");
  scanf("%d",&m);  
  switch (m){
    case 1:               /*判断能否为输入状态参数*/
         xstate = 0;
		 break;
    case  2:
         xstate = 1;
		 break;
    case  3:
          direction(m);
		  break;   /*判断选取数组a中的元素*/
    case  4:
          direction(m);
		  break;
    case  5:
    case 10:
          printf("\nplease input the distance:");
          scanf("%d",&n);
          move(n,xstate);
		  break;      /*用数组a中的返回值，对数组floor进行操作*/
    case  6:
          print();
		  break;     /*打印floor数组*/ 
    case  9:        
          wind = 0;
		  break;      /*退出程序标志*/
          }
    }     
    system("pause");
    return 0;
} 
int direction(int m1)
{
  if (m1 == 3){
             switch (state){
                     case "a":
                          Dstate[0][0] = 1;
                          Dstate[0][1] = 0;
                          state = "d";
                          break;
                           
                     case "b":
                          Dstate[0][0] = -1;
                          Dstate[0][1] = 0;
                          state = "c";
                          break;     
                     case "c":
                          Dstate[0][0] = 0;
                          Dstate[0][1] = 1;
                          state = "a";
                          break;
                     case "d":
                          Dstate[0][0] = 0;
                          Dstate[0][1] = -1;
                          state = "b";
                          break;
                          }
       }          
  else if(m1 == 4)
      {
             switch (state){
                    case "a":
                          Dstate[0][0] = -1;
                          Dstate[0][1] = 0;
                          state = "c";
                          break;
                    case "b":
                          Dstate[0][0] = 1;
                          Dstate[0][1] = 0;
                          state = "d";
                          break;
                    case "c":
                          Dstate[0][0] = 0;
                          Dstate[0][1] = -1;
                          state = "b";
                          break;
                    case "d":
                          Dstate[0][0] = 0;
                          Dstate[0][1] = 1;
                          state = "a";
                          break;
                         }
      }
    return 0;                
  }
int move (int n1,int x1)
  { 
     while(n1>0)
        {  i = i+ Dstate[0][0];
           j = j+ Dstate[0][1];
           floor[i][j] = x1;
           n1--;       
          } 
          return 0;                
  } 
  
 int print(void)
  { int i1=0,j1=0;
     for(i1=0;i1<50;i1++)
      {for(j1=0;j1<50;j1++)
        {  if (floor[i1][j1] == 1)
                printf("%c","*");
           else 
             printf("%d",floor[i1][j1]=0);
           }  
           printf("\n");
        }
    return 0;    
 }  

这段代码不是本人写的（本人不会）在这段代码中，本人有很多问题：1：首先是Dtate这个数组是用来干什么的？2：在direction这个函数中，实现上下左右的时候，应用Dstate这个数组来实现的，但是本人没有读懂是怎么使得上下左右走的；3:能不能帮本人讲讲move这个函数