如果输入无效输入（在 C 编程中），如何使该程序再次要求输入？ [复制]

Question

检查发生在两个不同的位置。

也就是说，在您输入a、s、m、d 或q 的位置以及输入第一个和第二个数字时。

在任何检查中，如果检查为假，它应该要求您重新输入您的输入。

我猜这可以通过在 while 循环检查中放置数字部分的 scanf 语句来完成，但是当我输入无效值（非数字）时，循环会无限运行。

所以我一定做错了什么。我已经使a、s、m、d 和q 部分大部分工作。

但第二部分似乎永远不会奏效。为此，我将失败的尝试留在了 while 循环中，而不是 //comments。

任何帮助将不胜感激！ 到目前为止，这是我的代码：

#include <stdio.h>
#include <ctype.h>

int main(void)
{
    char ch;
    float num1,num2,answer;
    printf("Enter the operation of your choice:\n");
    printf("a. add      s. subtract\n");
    printf("m. multiply q. divide\n");
    printf("q. quit\n");
    while ((ch = getchar())!='q')
    {
        printf("Enter the operation of your choice:\n");
        printf("a. add      s. subtract\n");
        printf("m. multiply q. divide\n");
        printf("q. quit\n");
        ch=tolower(ch);
        if (ch=='\n')
            continue;
        else
        {
            switch(ch)
            {
                case 'a':
                //The code below is what I have tried to make work.
                //This code would also be copy pasted to the other cases,
                //of course with the correct operations respectively being used.
                //
                //printf("Enter first number: ")
                //while(scanf("%f",&num1)==0)
                //{
                //  printf("Invalid input. Please enter a number.");
                //  scanf("%f",&num1);
                //}
                //printf("Enter second number: ")
                //while(scanf("%f",&num2)==0)
                //{
                //  printf("Invalid input. Please enter a number.");
                //  scanf("%f",&num2);
                //}
                //answer = num1 + num2;
                //printf("%f + %f = %f\n",num1,num2,answer);
                //break;
                //
                //I have also tried to make this work using do-while loops
                printf("Enter first number: ");
                scanf("%f",&num1);
                printf("Enter second number: ");
                scanf("%f",&num2);
                answer = num1 + num2;
                printf("%f + %f = %f\n",num1,num2,answer);
                break;
            case 's':
                printf("Enter first number: ");
                scanf("%f",&num1);
                printf("Enter second number: ");
                scanf("%f",&num2);
                answer = num1 - num2;
                printf("%f - %f = %f\n",num1,num2,answer);
                break;
            case 'm':
                printf("Enter first number: ");
                scanf("%f",&num1);
                printf("Enter second number: ");
                scanf("%f",&num2);
                answer = num1 * num2;
                printf("%f * %f = %f\n",num1,num2,answer);
                break;
            case 'd':
                printf("Enter first number: ");
                scanf("%f",&num1);
                printf("Enter second number: ");
                scanf("%f",&num2);
                answer = num1 / num2;
                printf("%f / %f = %f\n",num1,num2,answer);
                break;
            default:
                printf("That is not a valid operation.\n");
                break;
        }
    }
}
return 0;
}

再次感谢您的帮助！ 你会是一个救生员！ 干杯! -Will S.

编辑：我的代码可以工作了！这是最终的代码...

#include <stdio.h>
#include <ctype.h>

int main(void)
{
    char ch;
    float num1,num2,answer;
    printf("Enter the operation of your choice:\n");
    printf("a. add      s. subtract\n");
    printf("m. multiply q. divide\n");
    printf("q. quit\n");
    while ((ch = getchar())!='q')
    {
        ch=tolower(ch);
        //Ignore whitespace
        if (ch=='\n')
            continue;
        else
        {
            switch(ch)
            {
                //Addition part
                case 'a':
                    //First number
                    printf("Enter first number: ");
                    //Check to see if input is a number
                while (scanf("%f",&num1)==0)
                {
                    printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");
                    scanf("%*s");
                }
                //Second number
                printf("Enter second number: ");
                while (scanf("%f",&num2)==0)
                {
                    printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");
                    scanf("%*s");
                }
                //Do math for respective operation
                answer = num1 + num2;
                //Print out result
                printf("%.3f + %.3f = %.3f\n", num1,num2,answer);
                break;
            //Subtraction part
            case 's':
                printf("Enter first number: ");
                while (scanf("%f",&num1)==0)
                {
                    printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");
                    scanf("%*s");
                }
                printf("Enter second number: ");
                while (scanf("%f",&num2)==0)
                {
                    printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");
                    scanf("%*s");
                }
                answer = num1 - num2;
                printf("%.3f - %.3f = %.3f\n", num1,num2,answer);
                break;
            //Multiplication part
            case 'm':
                printf("Enter first number: ");
                while (scanf("%f",&num1)==0)
                {
                    printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");
                    scanf("%*s");
                }
                printf("Enter second number: ");
                while (scanf("%f",&num2)==0)
                {
                    printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");
                    scanf("%*s");
                }
                answer = num1 * num2;
                printf("%.3f * %.3f = %.3f\n", num1,num2,answer);
                break;
            //Division part
            case 'd':
                printf("Enter first number: ");
                while (scanf("%f",&num1)==0)
                {
                    printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");
                    scanf("%*s");
                }
                printf("Enter second number: ");
                while (scanf("%f",&num2)==0)
                {
                    printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");
                    scanf("%*s");
                }
                //Check for if number is a zero
                while (num2==0)
                {
                    printf("Please enter a non-zero number, such as 2.5, -1.78E8, or 3: ");
                    while (scanf("%f",&num2)==0)
                    {
                        printf("Invalid input. Please enter a number, such as 2.5, -1.78E8, or 3: ");
                        scanf("%*s");
                    }
                }
                answer = num1 / num2;
                printf("%.3f / %.3f = %.3f\n", num1,num2,answer);
                break;
            //For if a non-valid operation is entered
            default:
                printf("That is not a valid operation.\n");
                break;
        }
    }
    printf("Enter the operation of your choice:\n");
    printf("a. add      s. subtract\n");
    printf("m. multiply q. divide\n");
    printf("q. quit\n");
}
printf("Bye.\n");
return 0;

}

回想起来，我可能可以不用 if/else 语句。

Answer 1

您的代码存在多个问题。首先在这个循环中

1. 您在失败时接受了两次输入

   while(scanf("%f",&num1)==0)   //Taking Input Here Once
   {
       printf("Invalid input. Please enter a number.");
       scanf("%f",&num1);             //Again Taking input.
   }
   

   相反，您想要检查scanf() 的返回值，如果是0，您将再次执行循环，所以这是执行此操作的方法：

   int l = 0;
   
   while(l==0){  //Checking l, if it is zero or not, if zero running loop again.
       printf("Invalid input. Please enter a number.");
       l = scanf("%f",&num1);                   //Storing Return Value of scanf in l   
   }
   

2. 当程序遇到任何带有scanf("%f" , &num1) 或scanf("%f" , &num2) 的行时，它将跳过所有空格并等待下一次输入。如果输入与格式规范不匹配，则输入不会被消耗并保留在输入缓冲区中。

   int l = 0;
   
   while(l==0){  //Checking l 
   
        printf("Invalid input. Please enter a number.");
        l = scanf("%f",&num1);  //scanf will look at the buffer if the input
                                //does not match, it will not be consumed 
                                //and will remain in buffer.       
   }
   

   换句话说，不匹配的字符永远不会被读取。所以当你输入例如a 字符，您的代码将无限循环，因为scanf 在同一字符上继续失败。

3. 当程序执行其最后一次scanf("%f",&num2) 调用时，由于 enter 缓冲区中存在换行符\n，因此由于ch = getchar()，换行符\n 得到存储在ch 和以下if 条件满足并再次执行循环。

    if(ch =='\n')
        continue;
   

Answer 2

while(scanf("%f",&num1)==0)
{
     printf("Invalid input. Please enter a number.");
      scanf("%f",&num1);
 }

这个循环每次迭代扫描两个数字。这不是你想要的。输了第二个scanf。

您还应该检查 EOF 和错误。

int result;
while((result = scanf("%f",&num1))==0)
{
     printf("Invalid input. Please enter a number.");
}

if (result == EOF) .... report an error and exit ...