what does the line int *(*(x[3])())[5]; do in C? [closed]

问题

This is my first time asking a question on Stack Overflow, so please fell free to tell me if I did anything wrong or not specific enough. I've been programming microcontrollers in C for about 4 now. Some days ago I took part in an Electronics Competition. One of the many questions was what exactly the C codeline

int *(*(x[3])())[5];

does. This is the line as I remember it. It is possible that a bracket was at an other location, but I think this was the line.

My guess is that x is an array of functionpointers of which we take the fourth element and dereference it. This function is then executed without handing over parameters. The returnvalue seems to be a pointer to a pointer an array of which we dereference once in order to get the address of the first element. We then choose the 6th element of that array. I have no idea what the int is for though...

Thank you very much for answering my question and have a nice day.

回答1:

The way to read hairy declarations like this is to find the leftmost identifier and work your way out, remembering the following precedence rules:

*a[n]      -- a is an array of pointer
(*a)[n]    -- a is a pointer to an array
*f()       -- f is a function returning a pointer
(*f)()     -- f is a pointer to a function

Applying these rules, we get

        x             -- x is a
        x[3]          -- 3-element array of
       (x[3])()       -- function returning
      *(x[3])()       -- pointer to
     (*(x[3])())[5]   -- 5 element array of
    *(*(x[3])())[5]   -- pointer to
int *(*(x[3])())[5];  -- int

As written, this declaration isn't valid; you can't have an array of function type. I imagine it was supposed to be something like

int *(*(*x[3])())[5];

where x is an array of pointers to functions returning pointers to arrays of pointers to int.

回答2:

Use cdecl:

$ cdecl
cdecl> explain int *(*(x[3])())[5]
declare x as array 3 of function returning pointer to array 5 of pointer to int

It's what cdecl wrote above.

回答3:

When you cannot access a tool to translate a declaration for you, you can remember this threes rules of thumbs

There are three basic atoms (excluding type names)

Atom [n] translates to Array (with n elements) of.
Atom () translates to Function(s) returning.
Atom * translates to Pointer(s) to.
Start from the identifier moving right, whenever you encounter an atom, translate it.
A closed facing parenthesis¹ or the end of line reverse the direction of parsing.

¹_{This means ) when moving right, ( when moving left.}

A super pedantic translation of int *(*(x[3])())[5] is below.

Start from the identifier, x.
Translation: x is
Parsing: int *(*(-[3])())[5]
Direction: Right.
Found atom, [3].
Translation: x is an array (with 3 elements) of
Parsing: int *(*(---)())[5]
Direction: Right.
Found ), reverse direction.
Translation: x is an array (with 3 elements) of
Parsing: int *(*(----())[5]
Direction: Left.
Found (, reverse direction.
Translation: x is an array (with 3 elements) of
Parsing: int *(*-----())[5]
Direction: Right.
Found atom ().
Translation: x is an array (with 3 elements) of functions returning
Parsing: int *(*-------)[5]
Direction: Right.
Found ), reverse direction.
Translation: x is an array (with 3 elements) of functions returning
Parsing: int *(*--------[5]
Direction: Left.
Found atom * Translation: x is an array (with 3 elements) of functions returning a pointer to
Parsing: int *(---------[5]
Direction: Left.
Found (, reverse direction Translation: x is an array (with 3 elements) of functions returning a pointer to
Parsing: int *----------[5]
Direction: Right.
Found atom [5] Translation: x is an array (with 3 elements) of functions returning a pointer **to an array (with 5 elements) of **
Parsing: int *------------
Direction: Right.
End of line, reverse direction. Translation: *x is an array (with 3 elements) of functions returning a pointer to an array (with 5 elements) of *
Parsing: int *------------
Direction: Left.
Found atom *. Translation: *x is an array (with 3 elements) of functions returning a pointer to an array (with 5 elements) of * pointers to
Parsing: int -------------
Direction: Left.
Found type name int. Translation: x is an array (with 3 elements) of functions returning a pointer to an array (with 5 elements) of pointers to ints
Parsing: -----------------
Direction: Left.

Translation: x is an array (with 3 elements) of functions returning a pointer to an array (with 5 elements) of pointers to ints.

You can also experiment on removing parenthesis:

int **(x[3]())[5] translate to x is an array (with 3 elements) of functions returning an array (with 5 elements) of pointers to pointers to ints.

int **x[3]()[5] translate to x is an array (with 3 elements) of functions returning an array (with 5 elements) of pointers to pointers to ints.
Same as above!

Of course this is quite error prone, I kept cdecl handy in writing this answer!

来源：https://stackoverflow.com/questions/37364198/what-does-the-line-int-x35-do-in-c

标签

arrays

pointers

function-pointers