CODEWITHSUNDEEP
c
hareldo
hareldo

Reputation: 1

What does this C declaration int *(*table())[10] do?

I am trying to write C code to create a two-dimensional array of function pointers, where every function takes no arguments and returns int. I have already tried this code:

int *(*table())[10];

but it didn't work. Do you have any idea how can I fix it?

Upvotes: 0

Views: 309

Answers (2)

Lundin
Lundin

Reputation: 213799

Regarding what the declaration you came up with does, that's a pretty icky declaration. You can unwind it with the clockwise/spiral rule.

Basically the table() part means there's a a function. The () is obsolete C, meaning a function taking any parameter. You should be using (void) instead.

The int* (* ... ) [10] part means that you have a function returning an array pointer to an array int* [10]. Such an array pointer is normally declared as int* (*name)[10] when written outside a function return value.

So you have a function declaration, of a function returning an array pointer, to an array of 10 int*, the function is taking an obsolete style parameter list.

You can understand it better with this example:

typedef int* (*arrptr)[10]; // to illustrate, typedef an array pointer int* (*)[10]

typedef arrptr func_t (void); // typedef a function type returning said array pointer

void func (func_t* x){}
// dummy function to demonstrate type compatibility, funct_t* is a function pointer 


int *(*table(void))[10]; // function declaration

int main (void)
{

  func(table); // table is of same type as the function pointer func_t*

  return 0;
}

int *(*table(void))[10]
{
  return 0;
}

Please note that hiding arrays and pointers behind typedefs is normally bad practice.

Regarding how to get an array of function pointers, it is way more straight-forward than interpreting the above gibberish.

  • int* func (void) This is a function.
  • int* (*func) (void) This is a function pointer to such a function.
  • int* (*func[10][10]) (void) This is a 2D array of function pointers.

Or preferably:

  • typedef int* func_t (void) This is a function type.
  • func_t* fptr This is a function pointer to such a function.
  • func_t* fptr[10][10] This is a 2D array of function pointers.

Upvotes: 1

Angew is no longer proud of SO
Angew is no longer proud of SO

Reputation: 171127

C declarations are built such that typing the declarators as an expression yields the type. So, let's assume we already have such a 2D array a. How to get int out of it?

  1. Index once, getting a 1D array: a[10].
  2. Index again, getting a pointer to function: a[10][10]
  3. Dereference the pointer, getting a function: *a[10][10]
  4. Call the function, getting an int: (*a[10][10])()

So the final declaration is:

int (*a[10][10])(void);

Of course, as usual, it's much easier with typedefs:

typedef int Fun(void);
typedef Fun *FunPtr;

FunPtr a[10][10];

Upvotes: 3

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