Because scanf() needs pointers, and &a is the pointer to an integer.
In your example, you need a comma and an ampersand in front of the c.
int main(void)
{
int a;
char c;
char s[21];
scanf("%d",&a);
scanf("%c", &c); // Fixed!
scanf("%20s", s);
printf("a = %d, c = %c, s = <<%s>>\n", a, c, s);
}
If you were reading a string, you would pass the string - a pointer to char - to scanf().
