Why did these variable apparently change type? - c

Question

Python spoiled me and trying to wrap my mind around C now is being a bloodbath of stupid errors. This is one I can't quite understand.
I wanted the C equivalent of Python's os.path.split, but there's no exact equivalent. strsep looks similar enough, but needs some massaging to be used simply.
First off, I defined my path type: a string of given length.
#define MAX_PATH_LEN 200 /* sigh */
typedef char t_path[MAX_PATH_LEN];
Then I wrote some code that does the actual massaging, attempting to avoid side effects -- just to keep things fool proof.
typedef struct {
t_path next;
t_path remainder;
} t_path_next
t_path_next path_walk_into(t_path path) {
t_path_next output;
t_path my_next, my_remainder = "/";
strncpy(my_next, path, MAX_PATH_LEN);
strsep(&my_next, my_remainder);
output.remainder = my_remainder;
output.next = my_next;
return output;
}
gcc, however, is not impressed.
badp#delta:~/blah$ gcc path.c -Wall
path.c: In function ‘path_walk_into’:
path.c:39: warning: passing argument 1 of ‘strsep’ from incompatible pointer type
/usr/include/string.h:559: note: expected ‘char ** __restrict__’ but argument is of type ‘char (*)[200]’
path.c:41: error: incompatible types when assigning to type ‘t_path’ from type ‘char *’
path.c:42: error: incompatible types when assigning to type ‘t_path’ from type ‘char *’
I am baffled by the note -- how are char ** and char (*)[200] really different -- but the error is even weirder. I want to assign a variable I declared t_path in a field of type t_path, but I don't get to.
Why is that?
For anybody interest here's the correctly working version of the function:
t_path_next path_walk_into(t_path path) {
t_path_next output;
t_path my_path, delim = "/";
char* my_path_ptr = my_path;
strncpy(my_path, path, MAX_PATH_LEN);
strsep(&my_path_ptr, delim); //put a \0 on next slash and advance pointer there.
if (my_path_ptr == NULL) //no more slashes.
output.remainder[0] = 0;
else
strncpy(output.remainder, my_path_ptr, MAX_PATH_LEN);
strncpy(output.next, my_path, MAX_PATH_LEN);
return output;
}

Answer 1

The errors: You can't directly assign to an array, such as a string, in C. You need to copy char by char, or call str(n)cpy, which does it for you.

Answer 2

For the warning : you are probably already aware that array may decay to pointer. That is, for example, what makes an array acceptable as an argument to a function where a pointer is expected. In your case, what you have is a pointer to an array : there is no reason for such a thing to get converted to a pointer to pointer.
For the record, the C99 standard says (6.3.2.1/3) :
Except when it is the operand of the sizeof operator or the unary & operator, or is a
string literal used to initialize an array, an expression that has type ‘‘array of type’’ is
converted to an expression with type ‘‘pointer to type’’ that points to the initial element of
the array object and is not an lvalue.
You're in the context of a unary & : no conversion for you.
For the error : it has already been answered, but array assignment is not directly possible. You might want to use strcpy or strncpy.