发生了两个不同的斑点。
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时。
在任何检查中，如果支票是错误的，则应要求您重新输入输入。
我猜想这可以通过在a时循环检查中为数字部分放置scanf语句来完成，但是当我输入无效的值（非数字）时，循环会无限运行。
所以我必须做错了什么。我在大多数情况下制作了a，s，m，d和q零件工作。
，但第二部分似乎从来没有起作用。为此，我将失败的尝试段循环放出，而是在//comments中。
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. subtractn");
printf("m. multiply q. dividen");
printf("q. quitn");
while ((ch = getchar())!='q')
{
printf("Enter the operation of your choice:n");
printf("a. add s. subtractn");
printf("m. multiply q. dividen");
printf("q. quitn");
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 = %fn",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 = %fn",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 = %fn",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 = %fn",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 = %fn",num1,num2,answer);
break;
default:
printf("That is not a valid operation.n");
break;
}
}
}
return 0;
}
再次感谢您的帮助！你将是救生员！干杯! - 将会
编辑：我有代码工作！这是最终代码...
#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. subtractn");
printf("m. multiply q. dividen");
printf("q. quitn");
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 = %.3fn", 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 = %.3fn", 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 = %.3fn", 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 = %.3fn", 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. subtractn");
printf("m. multiply q. dividen");
printf("q. quitn");
}
printf("Bye.n");
return 0;
}
回头看，我可能可以没有/else语句。
您的代码有多个问题。首先在此循环中
-
您在失败上进行两次输入
while(scanf("%f",&num1)==0) //Taking Input Here Once { printf("Invalid input. Please enter a number."); scanf("%f",&num1); //Again Taking input. }相反，您想要的是检查
scanf()的返回值，如果是0，您将再次执行循环，因此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 } -
当程序与
scanf("%f" , &num1)或scanf("%f" , &num2)遇到任何线时，它将跳过所有白色空间并等待下一个输入。如果输入与格式规范不匹配，则输入不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读取。因此，当您键入例如一个 a 字符，当
scanf继续失败时，您的代码将无限期地循环。 -
当程序执行最后一个
scanf("%f",&num2)呼叫时，由于 enter hewlinen字符中存在于缓冲区中，因此由于ch = getchar()，新行n被存储在ch中，并且遵循if条件满足，循环再次执行。if(ch =='n') continue;
while(scanf("%f",&num1)==0)
{
printf("Invalid input. Please enter a number.");
scanf("%f",&num1);
}
此循环在迭代中扫描两个数字。这不是您想要的。丢失第二个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 ...