/* 对于下面提示的警告，网上有说 一般来说,字符里面比如'. '含有空格的话,会提示这个警告,经检查确认没有空格，不过警告依然继续，或者出现下面的页面提示。

*/
#include
#include
main()
{
      char c;
      int a,b;
      int n;
      printf("请输入你要运算的第一个数的值:");
      scanf("%d",a);
      printf("\n");
      printf("请输入你要运算的第二个数的值:");
      scanf("%d",b);
      printf("\n");
      printf("\n");
      printf(" 按&即将进入按位与，按|即将进入按位或，按^即将进入按位异或,按~即将进入求反操作，按<<即将进入左移，按>>即将进入右移操作\n");
      printf("请确认你要进行的运算：");
      scanf("%c",c);
      switch(c){
                case '&':printf("运算结果为：",a&b);break;
                case '|':printf("运算结果为：",a|b);break;
                case '^':printf("运算结果为：",a^b);break;
                case '~':printf("运算结果为:",~a,~b);break;
                case '<<':
                     printf("请确认你要左移的位数n:");
                     printf("\n");
                     printf("运算结果为：",a<<n,b<<n);
                     break;
                case '>>':
                     printf("请确认你要右移的位数n:");
                     printf("\n");
                     printf("运算结果为：",a>>n,b>>n);
                     break;
                default :printf("您的输入有误，请重新输入！");
                }
      system("pause");
      
}

[此贴子已经被作者于2015-11-30 11:22编辑过]

感谢批评、

是个问题。

/* 现在的运行时这样的

直接跳转到 default选项了，这是为什么？

*/

#include <stdio.h>
#include <math.h>
main()
{
      char c;
      int a,b;
      int n;
      printf("请输入你要运算的第一个数的值:");
      scanf("%d",&a);
      printf("\n");
      printf("请输入你要运算的第二个数的值:");
      scanf("%d",&b);
      printf("\n");
      printf("\n");
      printf(" 按&即将进入按位与，按|即将进入按位或，按^即将进入按位异或,按~即将进入求反操作，按<<即将进入左移，按>>即将进入右移操作");
      printf("\n");
      printf("请确认你要进行的运算：");
      scanf("%s",&c);
      switch('c'){
                case '&':printf("运算结果为：",a&b);break;
                case '|':printf("运算结果为：",a|b);break;
                case '^':printf("运算结果为：",a^b);break;
                case '~':printf("运算结果为:",~a,~b);break;
                case '<<':
                     printf("请确认你要左移的位数n:");
                     printf("\n");
                     printf("运算结果为：",a<<n,b<<n);
                     break;
                case '>>':
                     printf("请确认你要右移的位数n:");
                     printf("\n");
                     printf("运算结果为：",a>>n,b>>n);
                     break;
                default:printf("您的输入有误，请重新输入！");
                }
      system("pause");
      
}

嗯，网上确实有这样的答案，不过对于这样的情况，我尝试着用双引号去使用，然而却是失败的。

对于 case label does not reduce to an integer constant
解释为：案例标签不减少到一个整型常量  即是说 case接收的不能为字符串，应该为单个字符或单个字节常量。

是不是需要这样写：
int c;
// 1代表一种情况，2代表一种情况。。。。。这样写，然后再 case1： 后面的句子中加以实现。
switch(c)
{case 1:
 case 2:
 ...
 default:
}


不过这样有木有很生硬？

嗯，是这样的。

/*

感谢各位，已经实现了这个子程序。*/

#include <stdio.h>
#include <math.h>

void Bit_operation()
{
     int a,b;
     int d = 0;
     int n;
     printf("  1：与运算     2：或运算     3：按位与或   4：求反操作  \n");
     printf("  5：左移操作   6：右移操作   7：Exit\n");
     printf("请输入你的选择");
     scanf("%d",&n);
     printf("\n");
     switch(n){
                case 1:
                     printf("请输入你要运算的第一个数的值:");
                     scanf("%d",&a);
                     printf("\n");
                     printf("请输入你要运算的第二个数的值:");
                     scanf("%d",&b);
                     d = a&b;      
                     printf("运算结果为：%d\n",d);break;
                case 2:
                     printf("请输入你要运算的第一个数的值:");
                     scanf("%d",&a);
                     printf("\n");
                     printf("请输入你要运算的第二个数的值:");
                     scanf("%d",&b);
                     d = a|b;
                     printf("运算结果为：%d\n",d);break;  
                case 3:
                     printf("请输入你要运算的第一个数的值:");
                     scanf("%d",&a);
                     printf("\n");
                     printf("请输入你要运算的第二个数的值:");
                     scanf("%d",&b);
                     d = a^b;
                     printf("运算结果为：%d\n",d);break;
                case 4:
                     printf("请输入你要运算的数值:");
                     scanf("%d",&a);
                     printf("\n");
                     d = ~a;
                     printf("运算结果为：%d\n",d);break;
                case 5:
                     printf("请输入你要运算的数值:");
                     scanf("%d",&a);
                     printf("请确认你要左移的位数:");
                     scanf("%d",&n);
                     d = a<<n;
                     printf("运算结果为：%d\n",d);break;
                case 6:
                     printf("请输入你要运算的数值:");
                     scanf("%d",&a);
                     printf("请确认你要左移的位数:");
                     scanf("%d",&n);
                     d = a>>n;
                     printf("运算结果为：%d\n",d);break;      
                default:
                     printf("您的输入有误，请重新输入！");
                }
}
main()
{
      char c;
      int a,b;
      int n;
      printf("欢迎来到运算小世界！！！\n");
      printf("\n");
      Bit_operation(a,b,n);
      system("pause");
      
}