CODEWITHSUNDEEP
c
Doruk Ozcan
Doruk Ozcan

Reputation: 19

Scanf Function with Char is Being Skipped in C

I'm new at c programming and I tried to write this basic program below but it keeps skipping the second scanf function(scanf(" %c", &sign)) whatever I do. Do you have any idea to solve this problem? In addition, I tried to use getchar function but it didn't work too.

SS of the error

#include <stdio.h>

int main()
{
    int num1, num2, mult, sum, subt;
    char sign;

    printf("Enter two numbers: ");
    scanf("%d%d", &num1, &num2);

    printf("\nEnter operator type: ");
    scanf(" %c", &sign);

    if(sign == '*')
        {mult = num1 * num2;
        printf("\nResult is %d\n\n\a", &mult);}

    else if(sign == '+')
        {sum = num1 + num2;
        printf("\nResult is %d\n\n\a", &sum);}

    else if(sign == '-')
        {subt = num1 - num2;
        printf("\nResult is %d\n\n\a", &subt);}

    else
        printf("\nOperator is not valid, program has ended\n\n\a");

system("pause");

return 0;
}

Upvotes: 1

Views: 142

Answers (2)

Nasrat Takoor
Nasrat Takoor

Reputation: 354

The problem is in these lines

printf("\nResult is %d\n\n\a", &mult);

printf("\nResult is %d\n\n\a", &sum);

printf("\nResult is %d\n\n\a", &subt);

Take off the ampersand &. You don't want to print the memory addresses of the variables (which are the large, seemingly random numbers), you want the values stored in the variables.

Upvotes: 3

chux
chux

Reputation: 153368

The problem is not enabling all warnings.

A good well enabled compiler will warn about:

int subt;
printf("\nResult is %d\n\n\a", &subt);}

with something like

warning: format '%d' expects argument of type 'int', but argument 2 has type 'int *' [-Wformat=]

To not just solve the little problem, but get to higher level issues, the best things to learn here is to use available resources to help you write error-free code efficiently.

With gcc, consider:

gcc -pedantic -Wall -Wextra -Wconversion ..."

Upvotes: 0

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