What does an asterisk in empty parentheses mean?

Question

What does this c code do?

{
    int (*func)();
    func = (int (*)()) code;
    (int)(*func)();
}

Especially I confused about subj.

Answer 1

It's a cast to a function pointer.

The sequence int (*)() is for a function pointer that takes an indeterminate number of arguments, and returns an int. Wrapping it in parentheses like (int (*)()), when combined with an expression, is casting the result of the expression.

The code you present, with comments:

// Declare a variable `func` which is a pointer to a function
int (*func)();

// Cast the result of the expression `code` and assign it to the variable `func`
func = (int (*)()) code;

// Use the variable `func` to call the code, cast the result to `int` (redundant)
// The returned value is also discarded
(int)(*func)();

Answer 2

Remember to do a typecast, we use the following:

(type_to_cast) value;

when you want to cast some value to a certain type.

Also remember you define a function pointer as

return_type (*pointer_name) (data_types_of_parameters);

And the type of a function pointer is

return_type (*) (data_types_of_parameters)

Finally, you can call a function with its pointer as

(*func_pointer)(arguments);

So, with those 4 points in mind, you see that your C code:

First defines a funciton pointer func.

Second, casts code as a function pointer and assign its value to func

Third, calls the function pointed by func, and casts the value reutrned to int.

Answer 3

• int (*func)(); declares func as a pointer to a function that takes any number of parameters and and return int.

• In statement func = (int (*)()) code;, a cast is applied to code and then assign it to the function pointer func.

• (int)(*func)(); doesn't make much sense. Cast is not needed and it discards the return value. The call should be simply like

  int var = func();  
  

  or

  int var = (*func)();