返回函数指针的函数的 C 语法

Question

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.

Answer 1

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

Answer 2

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;

Answer 3

The same as with the typedef, only you place your function definition in place of its name.

Here's how f2 would look like:

typedef int (*(*f2)(double))(float);

You can do f3 as an exercise, since I'm assuming this is homework ;)

Answer 4

Just don't. It can be done, but it will be very confusing. Typedef's are there to ease writing and reading this short of code.

A function f that takes no arguments and returns a function pointer int (*)(float) would probably be something like (untested):

int (*f())(float);

Then for the rest you just need to keep adding parenthesis until it looks like lisp.

Answer 5

Learn the the right-left rule :

The "right-left" rule is a completely regular rule for deciphering C declarations. It can also be useful in creating them.

Answer 6

Use std::function :

typedef std::function<int(float)> f1;
typedef std::function<f1(double)> f2;
typedef std::function<f2(int)>    f3;

or

typedef std::function<std::function<std::function<int(float)>(double)>(int)> f3;

Answer 7

Inspired by John Bode's answer, I want to present it in a graphical way. Maybe it will help you to understand how the compiler would lex it into an AST: