Format '% s expects an argument of type char *, but argument 2 is of type' char (*) [64]

Question

I am creating a program using C, and I have this line in my code:

scanf("%s", &path);

When I compile the source file, I get this warning:

main.c:84:2: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘char (*)[64]’ [-Wformat]

And this is the declaration for the variable path:

char path[64];

Why am I seeing this error? And how can I solve it?

Answer 1

The array is already a pointer object (as indicated by a dream pointer). You do not need an operator &, since the announcement

char path[64];

equivalent to setting a pathpointer to a 64-byte memory area.

Answer 2

%s , char *, char. &path, . path , ( , &path[0]).

Answer 3

scanf("%s", path);, , path - , - (array == & array)