Answer:
Why does the call
char s[30];
scanf("%s", s);
work? I thought you always needed an & on each variable passed to scanf.
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.
char s[30];
scanf("%s", s);
work? I thought you always needed an & on each variable passed to scanf.
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.
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