C language yes or no if statement with a char?

Question

#include <stdio.h>
int main(void)
{
    char fever, cough; /*Sets the chars.*/

    printf("Are you running a fever? (y/n)\n"); /*Asks if they have a fever and saves their input.*/
    scanf("%c",&fever);

    printf("Do you have a runny nose/cough? (y/n)\n"); /*Asks if they have a cough and saves their input.*/
    scanf(" %c",&cough);

    printf("Please verify the folling information.\nFever: %c \nRunny nose/cough: %c \n",fever,cough); /*Asks if the following info is correct.*/


    if ((fever=y) && (cough=y))
    printf("Your recommendation is to see a doctor.");

    else if ((fever=n) && (cough=y))
    printf("Your recommendation is to get some rest.");

    else if ((fever=y) && (cough=n)
    printf("Your recommendation is to see a doctor.");

    else
    printf("Your are healthy.");
return 0;
}

I get errors for the y's and n's

Answer 1

(fever=y) is assignment.

You need (fever == 'y')

Note the ' (quotes) and also, the conditional check == instead of =

This needs to be fixed in every occurance.

if ((fever == 'y') && (cough == 'y')) {
    printf("Your recommendation is to see a doctor.");
}

else if ((fever == 'n') && (cough == 'y')) {
    printf("Your recommendation is to get some rest.");
}

else if ((fever == 'y') && (cough == 'n') {
    printf("Your recommendation is to see a doctor.");
}

Answer 2

1 - in C programming language and many others equal ( = ) mean assignment operator. it means you give value to variable then if you need to say fever is equal to y you have to use == ,double equal mean equal.

2 - when you wanna say var equal to character ,you must write character in quotes like 'y'.

here you can see the correct way :

if ((fever == 'y') && (cough == 'y'))
printf("Your recommendation is to see a doctor.");

Answer 3

In order for the second scanf to read an input you must put a space before both %c scanf lines in the code.

Looking like this scanf(" %c", &fever) and scanf(" %c, &cough) the reason being that when you press enter after your first input of y or n the compiler reads this as an input itself. I was told this is a tricky part of scanf.

Answer 4

There are some mistakes in your code .

I - The variable fever & cough have character datatypes so in if condition the comparison variable must be in single quotes like 'y' & 'n' .

II - In the if condition you used the assignment operator '=' .This is wrong logic here . you must use the comparison operator '==' .

#include <stdio.h>
int main(void)
{
    char fever, cough; /*Sets the chars.*/

    printf("Are you running a fever? (y/n)\n"); /*Asks if they have a fever and saves their input.*/
    scanf("%c",&fever);

    printf("Do you have a runny nose/cough? (y/n)\n"); /*Asks if they have a cough and saves their input.*/
    scanf(" %c",&cough);

    printf("Please verify the folling information.\nFever: %c \nRunny nose/cough: %c \n",fever,cough); /*Asks if the following info is correct.*/


    if ((fever=='y') && (cough=='y'))
    printf("Your recommendation is to see a doctor.");

    else if ((fever=='n') && (cough=='y'))
    printf("Your recommendation is to get some rest.");

    else if ((fever=='y') && (cough=='n')
    printf("Your recommendation is to see a doctor.");

    else
    printf("Your are healthy.");
return 0;
}

Answer 5

Always remember when using string you need to use "string"(" ") for character use 'character'(' ').

Example,
char *s = "Hi i am fine"; //string
char *c = 'g'; //character

Make the changes accordingly in your code, to check for character 'y' and 'n'