如何检查变量是否不正确

Question

Basically I want to check if char a is not 'y' or 'n' . Been trying to figure it out for an hour now and could not find anything.

#include<stdio.h>
int yesno(char a){
    do{
        printf(":");
        scanf("%s",&a);
        if((a!='y')||(a!='n')){
            printf("Incorrect awnser, try again\n");
        }
    }while((a!='y')||(a!='n'));
}
int main(){
    printf("************************************\n");
    printf("*Welcome to 'noname' 0.01          *\n");
    printf("*Do you want to start y/n?         *\n");
    printf("************************************\n");
    yesno(1);
return 0;
}

Could someone tell me how to do that? How to make it check if something is NOT something. This code is what my understanding of C allows me to create, but it's not working correctly, it just loops: Incorrect answer, try again

Answer 1

"if((a!='y')||(a!='n')){
    printf("Incorrect awnser, try again\n");
}"

You could use a switch-case statement, with the default being the printf() statement.

switch(a) {
    case 'y':
    // do nothing
    break;
    case 'n':
    // do nothing
    break;
    default:
    printf("Incorrect answer, try again\n");
};

Answer 2

As suggested in the comments, you should use the && operator instead of || to check if a is neither 'y' nor 'n' like

    if((a!='y') && (a!='n')){
        printf("Incorrect awnser, try again\n");
    }
}while((a!='y') && (a!='n'));

The condition ((a!='y') || (a!='n')) will always be true as a cannot be 'y' and 'n' at the same time.

And in yesno() , a is a char and not a char array capable of holding strings.

So

scanf("%s",&a);

should be

scanf("%c",&a);

as the format specifier for a char is %c whereas %s is for strings.

Even if a were a char array to store a string, the scanf() should've been

scanf("%s",a);

as array name in C decay into pointer to its base element.

---

And you could do away with that argument of yesno() as the initial value of a is irrelevant since a new value of a is read in at the beginning of yesno() and any changes made to a will not have any repercussion back in main() as the parameter is passed by value and yesno() doesn't return a value.

So, instead of making a a parameter, you could make it a local variable of yesno() like

void yesno(){//or yesno(void)
    char a;
    do
    {
        printf(":");
        scanf("%c",a);
        if((a!='y') && (a!='n')){
            printf("Incorrect awnser, try again\n");
        }
    }while((a!='y') && (a!='n'));
}

And call it like

yesno();

in main() .

Answer 3

You can use continue and break statements inside the if condition.

#include <stdio.h>

int yesno(char a){

    do{

        printf(":");
        scanf("%s",&a);
        if((a=='y')||(a=='n')){
            break;
        }
        else{
            printf("Incorrect answer, try again\n");
            continue;           
        }

    }while((a!='y')||(a!='n'));
}

int main(){

    printf("************************************\n");
    printf("*Welcome to 'noname' 0.01          *\n");
    printf("*Do you want to start y/n?         *\n");
    printf("************************************\n");
    yesno(1);

return 0;
}

I just changed the if condition and added the else part.