C syntax for functions returning function pointers
Consider the following typedefs :
typedef int (*f1)(float);
typedef f1 (*f2)(double);
typedef f2 (*f3)(int);
f2
is a function that returns a function pointer. The same with f3
, but the type of the function, the pointer to which f3
returns, is f2
. How can I define f3
without the typedefs? I know typedefs are the cleaner and easier to understand way to define f3
. However, my intention here is to understand C syntax better.
Start with your declaration for f1
:
int (*f1)(float);
You want f2
to be a pointer to a function returning f1
, so substitute f1
in the declaration above with the declaration for f2
:
int (* f1 )(float);
|
+-----+-----+
| |
v v
int (*(*f2)(double))(float);
The declaration reads as
f2 -- f2
*f2 -- is a pointer
(*f2)( ) -- to a function
(*f2)(double) -- taking a double parameter
*(*f2)(double) -- returning a pointer
(*(*f2)(double))( ) -- to a function
(*(*f2)(double))(float) -- taking a float parameter
int (*(*f2)(double))(float) -- returning int
You repeat the process for f3
:
int (*(* f2 )(double))(float);
|
+---+----+
| |
v v
int (*(*(*f3)(int))(double))(float);
which reads as
f3 -- f3
*f3 -- is a pointer
(*f3)( ) -- to a function
(*f3)(int) -- taking an int parameter
*(*f3)(int) -- returning a pointer
(*(*f3)(int))( ) -- to a function
(*(*f3)(int))(double) -- taking a double parameter
*(*(*f3)(int))(double) -- returning a pointer
(*(*(*f3)(int))(double))( ) -- to a function
(*(*(*f3)(int))(double))(float) -- taking a float parameter
int (*(*(*f3)(int))(double))(float); -- returning int
In C++, the miracle of templates can make this a tad easier.
#include <type_traits>
std::add_pointer<
std::add_pointer<
std::add_pointer<
int(float)
>::type(double)
>::type(int)
>::type wow;