检查发生在两个不同的位置。
也就是说，您输入a，s，m，d或q以及输入第一个和第二个数字的位置。
在任何检查中，如果检查结果为false，则应要求您重新输入输入。
我猜这可以通过在while循环检查中为数字部分放置scanf语句来完成，但是当我输入无效值（非数字）时，循环无限运行。
所以我一定做错了。我已将a，s，m，d和q部分用于大部分工作。
但第二部分似乎永远不会起作用。为此，我将失败的尝试留在while循环中，而是放在//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！ 雅将是一个救命的人！ 干杯! - 会不会。
编辑：我让我的代码工作了！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;
}
回顾它，我可能没有if / else语句。
答案 0 :(得分：2)
scanf此循环每次迭代扫描两个数字。这不是你想要的。丢失第二个int result;
while((result = scanf("%f",&num1))==0)
{
printf("Invalid input. Please enter a number.");
}
if (result == EOF) .... report an error and exit ...
。
您还应检查EOF和错误。
{{1}}
答案 1 :(得分：2)
您的代码存在多个问题。首先是在这个循环中
您在失败时输入两次
while(scanf("%f",&num1)==0) //Taking Input Here Once
{
printf("Invalid input. Please enter a number.");
scanf("%f",&num1); //Again Taking input.
}
相反，您想要的是检查scanf()的返回值，如果它是0，您将再次执行循环，所以这将是这样做的方法：
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
}
当程序遇到scanf("%f" , &num1)或scanf("%f" , &num2)的任何行时，它将跳过所有空格并等待下一个输入。在输入与格式规范不匹配的情况下，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，永远不会读取不匹配的字符。所以当你输入例如一个 a 字符，当scanf继续在同一个字符上失败时，您的代码将无限循环。
当程序执行上一次scanf("%f",&num2)调用时，由于输入，缓冲区中存在换行符\n，因此{{1 }，新行ch = getchar()存储在\n中，后面的ch条件满足，循环再次执行。
if