[FIXED] Why in C does taking input using int format to a char change the value of another char variable?

Issue

I took input in "%d" format into a character using scanf() because I didn't want more than 8 bits. But doing so changed the value of another char variable.

Code:

#include <stdio.h>
int main() {
        char a;
        char b;
        printf("Enter a: ");
        scanf("%c", &a);
        printf("a = %c\n", a);
        printf("Enter b: ");
        scanf("%d", &b);
        printf("\na = %d\n", a);
        printf("b = %d\n", b);
}

Output:

Enter a: c
a = c
Enter b: 56

a = 0
b = 56

Screenshot

Solution

scanf reads data from stdin and stores them according to the parameter format into the locations pointed by the additional arguments.

scanf("%d", &b);

With %d you are storing an integer into the char variable that can not hold an integer. the Variable now grows into the other variable in stack above it.

if you compile with the flag "-fstack-protector-all" you should get the *** stack smashing detected *** error on execution.

Answered By - JBahn77

Answer Checked By - Katrina (PHPFixing Volunteer)