Why print 8 value when input char for %d

Question

#include <stdio.h>

int main() {
int a, b;
printf("Enter two numbers:");
scanf("%d %d", &a, &b);

printf("\nSum: %d", a+b);

return 0;
}
Input - c h
Output - 8

Answer 1

That's a garbage value. a and b are integers and scanf("%d") will look for integers, not characters. If you enter characters when the program expects integers, it's like you enter nothing at all. So it will just print the sum of the two values that were in memory where a and b are stored now (they are garbage values since you didn't initialize them)