Q: Why does the call
char s[30];
scanf("%s", s);
work?
I thought you always needed an &
on each variable passed to scanf.
A: You always need a pointer; you don't necessarily need an explicit &. When you pass an array to scanf, you do not need the &, because arrays are always passed to functions as pointers, whether you use & or not. See questions 6.3 and 6.4. (If you did use an explicit &, you'd get the wrong type of pointer; see question 6.12.)
Hosted by Eskimo North