Why does the call char scanf work?

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Why does the call
char s[30];
scanf("%s", s);

work? I thought you always needed an & on each variable passed to scanf.

✍: Guest

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.

2015-10-30, 1096👍, 0💬

💬 2021-01-21 John Doe: Hello

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Why does the call char scanf work?
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