问题
For practice, I'm trying to :
Declare fubar to be a pointer to a function that takes a pointer to a char and returns a pointer to an array of 24 elements where each element is a pointer to a struct foo.
My logic is:
-fubar is a pointer to a function taking a char pointer:
(*fubar)(char*)
-...returning a pointer to an array of 24 elems of where each elem is a struct foo:
(struct foo *)(*fubar)(char*)[24]
Is my logic correct?
回答1:
After fixing the syntax error and removing the parentheses around struct foo *,
struct foo* (*fubar)(char*)[24]
...the one part that you got wrong is that it actually returns an array of pointers, not a pointer to an array. In order to declare a pointer to the array, you need an extra set of parentheses:
struct foo (*(*fubar)(char*))[24]
You can pretend that that the star belongs to an identifier (i.e., the name of the array) inside the parentheses.
回答2:
Functions never return arrays, they may return a pointer (conventionally to the first cell of an array), since as return value or as argument an array decays to a pointer.
So, declare with a typedef the signature of your function:
typedef struct foo** funsig_t(char*);
notice that if you omit the typedef you would declare a function funsig_t of the desired signature.
then declare a pointer using that typedef:
funsig_t* fubar;
来源：https://stackoverflow.com/questions/27122155/declaring-a-function-pointer-returning-an-array