<分区>
检查发生在两个不同的地方。
也就是说，您在其中输入 a、s、m、d 或 q 以及当您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 循环检查中为数字部分放置一个 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。
所以我一定是做错了什么。我制作了 a、s、m、d 和 q 部分大部分时间都在工作。
但是第二部分似乎从来没有奏效。为此，我将失败的尝试留在了 while 循环中，而不是在 //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. 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! 你会是一个救命恩人! 干杯! -威尔·S。
编辑:我的代码可以工作了!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. 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;
回过头来看，我可能可以不用 if/else 语句。