C function declaration (* (void (*) () 0 )();

When I look at some excellent open-source projects, I can always see various pointers and usage of various functions. I still have an obscure understanding of this aspect. today I will take the time to complement it.

When I look at some excellent open-source projects, I can always see various pointers and usage of various functions. I still have an obscure understanding of this. today I will take the time to complement it,

After turning over C language traps and defects, the following is a brief summary:

Let's take a look at a statement:

(* (Void (*) () 0 )();

We can only use metamorphosis to describe it. next we will analyze it step by step:

The simplest way to declare variables:

Int a; // declare a variable of the int type

Simplest function declaration method

Int fn (); // declares that the result type returned by the fn function is int.

The simplest method of pointer declaration:

Int * p = 0; // declare a pointer to the int type p, initialized to 0.

The function that returns the pointer:

Int * ();

The simplest method to declare function pointers:
Int (* h )();

H is a function pointer. what should I do if I want to forcibly convert the type?

In fact, it is very easy.

Simple type conversion:

12	Void * p = 10; printf ("% d \ n", (int *) p );

The function pointer is the same,

For example, what if we want to forcibly convert a function to an int (* h) () type function pointer?

You only need to (int (*) (expression...

What if I want to declare a (int (*) () type pointer function?

Step 1 declare a function pointer

Int (* )();

You only need to use (int (*) () as a type to forcibly convert.
(* (Int (*) () )()

It is actually a function pointer of the type (int ..