#include <stdio.h>
int main()
{
int a,b;
printf("enter number\n");
scanf("%d",&a);
printf("enter number\n");
scanf("%d",&b);
printf("Numbera are %d %d",a,b);
return 0;
}
Running the Program
Input 12 a
Outptut 12 0
Should not I get output 12 97(ASCII value of a)
No, you should not get "12 97". That's not the way %d works. %d expects to see digits, and digits only.
If you modify your program like this:
int r1, r2;
printf("enter number\n");
r1 = scanf("%d",&a);
printf("enter number\n");
r2 = scanf("%d",&b);
printf("scanfs returned %d %d\n", r1, r2);
and type "12 a" again, you will see
scanfs returned 1 0
indicating that the second scanf call did not successfully read or convert anything. (When you use scanf, it's always a good idea to check the return value to make sure all your conversions worked.)
If you want to read a character as a character, you can use %c:
char b;
r2 = scanf(" %c",&b);
Beware, though, that %c is a little different from most of the other scanf format specifiers. You may need an explicit leading space, as I have shown here. See this other question for more information. (Thanks to @user202729 for the link.)
You cannot enter a character when scanning for a number and have it automatically translated. You would need to read it into a char and convert to int.
