C: can't figure out what is causing an infinite loop - c

Question

I'm a beginner, and as an exercise I have to code a very simple calculator that modifies the stack by the number and the operator the user inputs.
This is the code:
#include <stdio.h>
int main (void)
{
long double x, stack = 0;
char op;
printf("Input an operator and a number:\n");
while ( op != 'q' )
{
scanf("%Lf %c", &x, &op);
switch (op)
{
case '+':
stack += x;
printf("= %Lg", stack);
printf("\n");
break;
case '-':
stack -= x;
printf("= %Lg", stack);
printf("\n");
break;
case '*':
stack *= x;
printf("= %Lg", stack);
printf("\n");
break;
case '/':
if (x == 0)
{
printf("Can't divide by 0.");
printf("\n");
break;
}
stack /= x;
printf("= %Lg", stack);
printf("\n");
break;
case 's':
stack = x;
printf("stack set to %Lg", x);
printf("\n");
break;
case 'q':
printf("Bye!\n");
break;
default:
printf("Unknown operator.\n");
break;
}
}
printf("Bye!\n");
return 0;
}
Now the problem is that whenever x is not an number, the program keeps looping. Why?
And how could I prevent the user from inputing anything but a number for x? If x was a char I'd use isdigit(), but that's not the case.
(By the way, I want x to be a long double so I can input numbers with decimals).

Answer 1

scanf() will convert as many values from the string as it can. If one cannot be converted, it stops processing the string. Either way, the function returns a value indicating how many values were converted.
If the first value (x) is not a number and cannot be converted, it stops. And so the second value (op) does not change. That causes your loop to continue.
Check the value returned by scanf() to confirm this behavior yourself.
Also, "I want x to be a long double so I can input numbers with decimals" - Values of type float and double also support numbers with decimals.
Finally, as others have point out, you should initialize op at the start of your code before testing if it equals "q". This will ensure the value is what you expect it to be at that time.