关于#C语言#的问题，如何解决？

C语言输入问题
问题：为什么我输入数字后程序没有继续往下执行，而是重复输入

#include 
int main()
{    
    printf("Please switch the number:\n");
    printf("1,Single \t\t\t\t2,Householeder\n3,Married(co-ownership) \t\t4,Married(divorce)\n");
    int num1;
    double salary;
    
    while (scanf("%d", &num1)==1);
    {    
        
        salary = 0;
        printf("Please enter your salary:");
        scanf("%lf", &salary);
        switch (num1)
        {
        case 1:if (salary >= 17850) printf("You need to pay tax $%lf.", 0.15 * 17850 + 0.28 * (salary - 17850));
              else printf("You need to pay tax $%lf.", 0.15 * salary); break;
        case 2:if (salary >= 23900) printf("You need to pay tax $%lf.", 0.15 * 23900 + 0.28 * (salary - 23900));
              else printf("You need to pay tax $%lf.", 0.15 * salary); break;
        case 3:if (salary >= 29750) printf("You need to pay tax $%lf.", 0.15 * 29750 + 0.28 * (salary - 29750));
              else printf("You need to pay tax $%lf.", 0.15 * salary); break;
        case 4:if (salary >= 14875) printf("You need to pay tax $%lf.", 0.15 * 14875 + 0.28 * (salary - 14875));
              else printf("You need to pay tax $%lf.", 0.15 * salary); break;
        }
        
    }

    return 0;
}

第九行你多写了一个分号
造成了一直循环输入

• while() { } 这是带有循环体的循环语句
• while(); 这是不带循环体的，只会一直执行该条语句

  #include <stdio.h>
  int main()
  {    
    printf("Please switch the number:\n");
    printf("1,Single \t\t\t\t2,Householeder\n3,Married(co-ownership) \t\t4,Married(divorce)\n");
    int num1;
    double salary;
    
    while (scanf("%d", &num1)==1)
    {    
        
        salary = 0;
        printf("Please enter your salary:");
        scanf("%lf", &salary);
        switch (num1)
        {
        case 1:if (salary >= 17850) printf("You need to pay tax $%lf.", 0.15 * 17850 + 0.28 * (salary - 17850));
              else printf("You need to pay tax $%lf.", 0.15 * salary); break;
        case 2:if (salary >= 23900) printf("You need to pay tax $%lf.", 0.15 * 23900 + 0.28 * (salary - 23900));
              else printf("You need to pay tax $%lf.", 0.15 * salary); break;
        case 3:if (salary >= 29750) printf("You need to pay tax $%lf.", 0.15 * 29750 + 0.28 * (salary - 29750));
              else printf("You need to pay tax $%lf.", 0.15 * salary); break;
        case 4:if (salary >= 14875) printf("You need to pay tax $%lf.", 0.15 * 14875 + 0.28 * (salary - 14875));
              else printf("You need to pay tax $%lf.", 0.15 * salary); break;
        }
        
    }
  
    return 0;
  }
  

望采纳！！！
因为你的程序中的 while (scanf("%d", &num1)==1); 后面的花括号是多余的，而且在 while 循环中没有任何逻辑来更新循环条件，因此这个循环会一直重复。

试着把 while (scanf("%d", &num1)==1); 删掉，或者把它改成 while (scanf("%d", &num1)==1) {...} 。这样程序就只会在第一次输入时执行一次了。

你可以在switch语句后面加上一个break语句来结束循环。

• 这有个类似的问题, 你可以参考下: https://ask.csdn.net/questions/7634235
• 这篇博客你也可以参考下：c语言解决分鱼问题
• 这篇博客也不错, 你可以看下c语言解决分鱼问题

望采纳！！！点击回答右侧采纳即可！！
这是因为在while循环中，没有添加循环体，因此程序会一直重复输入，而不会继续往下执行。要让程序继续往下执行，需要在while循环中添加循环体，如下所示：

#include <stdio.h>
int main()
{    
    printf("Please switch the number:\n");
    printf("1,Single \t\t\t\t2,Householeder\n3,Married(co-ownership) \t\t4,Married(divorce)\n");
    int num1;
    double salary;
    
    while (scanf("%!d(MISSING)", &num1)==1)
    {    
        salary = 0;
        printf("Please enter your salary:");
        scanf("%!l(MISSING)f", &salary);
        switch (num1)
        {
        case 1:if (salary >= 17850) printf("You need to pay tax $%!l(MISSING)f.", 0.15 * 17850 + 0.28 * (salary - 17850));
              else printf("You need to pay tax $%!l(MISSING)f.", 0.15 * salary); break;
        case 2:if (salary >= 23900) printf("You need to pay tax $%!l(MISSING)f.", 0.15 * 23900 + 0.28 * (salary - 23900));
              else printf("You need to pay tax $%!l(MISSING)f.", 0.15 * salary); break;
        case 3:if (salary >= 29750) printf("You need to pay tax $%!l(MISSING)f.", 0.15 * 29750 + 0.28 * (salary - 29750));
              else printf("You need to pay tax $%!l(MISSING)f.", 0.15 * salary); break;
        case 4:if (salary >= 14875) printf("You need to pay tax $%!l(MISSING)f.", 0.15 * 14875 + 0.28 * (salary - 14875));
              else printf("You need to pay tax $%!l(MISSING)f.", 0.15 * salary); break;
        }
        
    }
    return 0;
}