这是我的程序#include "stdio.h"
void main()
{int number; char today[20];
    printf("please put the number:");
     scanf("%d",&number);
     switch(number)
     {case  1: today="monday";break;
      case  2: today="tuseday";break;
      case  3: today= "wendnesday";break;
      case  4: today="thursday";break;
      case  5: today="friday";break;
      case  6: today="saturday";break;
      case  7: today="sunday";break;
      default: today="no this day";};
     printf("%s",today);
   
}显示有错误

这是别人帮我改的#include "stdio.h"
void main()
{int number; char today[20],*p=today;
     printf("please put the number:");
      scanf("%d",&number);
      switch(number)
      {case   1:p="monday";break;
       case   2:p="tuseday";break;
       case   3:p="wendnesday";break;
       case   4:p="thursday";break;
       case   5:p="friday";break;
       case   6:p="saturday";break;
       case   7:p="sunday";break;
       default:p="no this day";};
       printf("%s\n",p);
     
}是正确的 额 我想问 为什么需要*p=taday 这一步 为什么这里的p不能直接用today代替

以下代码在Dev C++中调试通过！何必用*p呢？

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main()
{    

      int number;

      char today[20];
      printf("please put the number:");
      scanf("%d",&number);
      switch(number)
      {case 1:
              strcpy(today,"monday");
              break;
       case 2:
              strcpy(today,"tuseday");
              break;
       case 3:
              strcpy(today,"wendnesday");
              break;
       case 4:
              strcpy(today,"thursday");
              break;
       case 5:
              strcpy(today,"friday");
              break;
       case 6:
              strcpy(today,"saturday");
              break;
       case 7:
              strcpy(today,"sunday");
              break;
       default:
              strcpy(today,"no this day");
       };
      

       printf("%s\n",today);
      

       system("PAUSE");
       return 0;
}

数组与指针的关系，回去看书吧