Array as function with arguments?

Question

I'm trying to learn c, so I tried reading some source code.
But I have no idea what this might mean:

static const char*(*const functab[])(void)={
        ram,date
};

The first part, static const char* is fine, as it seems to be a function (has an argument of type void), static should mean that it is only visible in this file and const char* should mean that the value cannot be changed but the address can be changed.
But in that case, it doesn't make sense after the last part following the function name, as it was the case with

static const char * date(void);
static const char * ram(void);

Instead of the function name there is (*const functab[]), a const array called functab containing addresses?
Is this some kind of wrapping function containing the functions ram and date? Some alternative way of declaring arrays?

Answer 1

functab is an array of function pointers (array of const function pointers, to be exact), which returns const char* and accepts no arguments.

Later,

 ... = { ram, date };

is a brace-enclosed initializer list which serves as the initializer for the array.

Answer 2

This is the way to define array of function pointers in C. So instead of calling function as ram(), using this array you can call it by (* functab[1]).

Below discussion has good examples of array of function pointer: How can I use an array of function pointers?