How do I stop making this C code loop? [duplicate] - c

Question

This question already has answers here:
Why is scanf() causing infinite loop in this code?
(16 answers)
Closed 6 years ago.
I have written some C code where you can pick from an answer by inputting a 1 or 2, and if you input a higher number it will bring you back to pick another number just fine. However, if I try to input something that's not a value or a number, like a string or character, the error message at the bottom will repeat infinitely. How can I make my code act the same as if you input a higher number than 1 or 2 when you input any other character? Here the code I use abstracted:
#include <stdio.h>
int a;
int main(){
b:scanf("%d", &a);
if(a==1)
{
a=0;
}
if(a==2)
{
a=0;
}
else
{
a=0;
printf("\nERROR: Please try again.\n\n");
goto b;
}
}
EDIT: Apparently the return value is still stuck in scanf() when it returns to it. How can I clear out scanf() of its return value?

Answer 1

Don't use gotos at all. Instead use while loops:
#include <stdio.h>
int main(void) {
int a, end = 1; //end determines if the loop should end
do { //a do-while loop - it's the same as a while loop, except it runs atleast once
scanf("%d", &a);
switch (a) { //switches the value of a
case 1:
case 2: printf("You entered %d\n", a);
end = 0; //sets end to 0, which will end the loop(see below)
break;
default: printf("\nERROR: Please try again.\n\n");
}
} while (end); //every non-zero value is true, so when I set end to 0, it will end the loop
return 0; //don't forget the return 0: it shows you that your program ran without error
}
So I wrote it that it ends as soon as you type a valid input. You also don't need to set a to zero, as you read it in again every time you run the loop.
EDIT: if you want to check for invalid input such as 5x, you can use the following:
int check, var, error;
char ch;
do {
error = 0;
check = scanf("%d%c", &var, &ch);
if (check != 2 || ch != '\n') {
printf("Wrong input. Try again -> ");
error = 1;
fflush(stdin);
}
} while (error);

Answer 2

#include <stdio.h>
int a;
int isNumeric(const char *str)
{
while(*str != '\0')
{
if(*str < '0' || *str > '9')
return 0;
str++;
}
return 1;
}
int main(){
char inputStr[10];
while(1){
scanf("%9s",inputStr);
if(!isNumeric(inputStr)){
a=0;
printf("\nERROR Not a number: Please try again.\n\n");
}else {
a = atoi(inputStr);
if(a==1){
a = 0;
}else if(a == 2){
a == 0;
}else{
a=0;
printf("\nERROR : Please try again.\n\n");
}
}`enter code here`
}
}
Have not tested. But I guess you will get a good idea. check strtol function. That is also useful.

Answer 3

Something like...
Note: Obviously 999 is an arbitrary value. Just chosen to give you an example.
#include <stdio.h>
int main(){
int a = 1;
while (a != 999){
scanf("%d", &a);
if(a==1)
{
a=0;
}
if(a==2)
{
a=0;
}
else if (a != 999)
{
a=0;
printf("\nERROR: Please try again.\n\n");
}
} // while()
} // main()