Except when it is the operand of the sizeof
, _Alignof
, or unary &
operators, or is a string literal being used to initialize an array in a declaration, an expression of type "N-element array of T
" will be converted ("decay") to an expression of type "pointer to T
", and it will evaluate to the address of the first element in the array.
The array me
is declared as a 20-element array of char
; normally, when the expression me
appears in your code, it will be treated as an expression of type "pointer to char
". If you had written
scanf("%s", me);
then you wouldn't have gotten the error; the expression me
would have been converted to an expression of the correct type.
By using the &
operator, however, you've bypassed that rule; instead of a pointer to char
, you're passing a pointer to an array of char
(char (*)[20]
), which is not what scanf
expects for the %s
conversion specifier, hence the diagnostic.