What's the syntax for declaring an array of function pointers without using a separate typedef?

问题

Arrays of function pointers can be created like so:

typedef void(*FunctionPointer)();
FunctionPointer functionPointers[] = {/* Stuff here */};

What is the syntax for creating a function pointer array without using the typedef?

回答1:

arr    //arr 
arr [] //is an array (so index it)
* arr [] //of pointers (so dereference them)
(* arr [])() //to functions taking nothing (so call them with ())
void (* arr [])() //returning void 

so your answer is

void (* arr [])() = {};

But naturally, this is a bad practice, just use typedefs :)

Extra: Wonder how to declare an array of 3 pointers to functions taking int and returning a pointer to an array of 4 pointers to functions taking double and returning char? (how cool is that, huh? :))

arr //arr
arr [3] //is an array of 3 (index it)
* arr [3] //pointers
(* arr [3])(int) //to functions taking int (call it) and
*(* arr [3])(int) //returning a pointer (dereference it)
(*(* arr [3])(int))[4] //to an array of 4
*(*(* arr [3])(int))[4] //pointers
(*(*(* arr [3])(int))[4])(double) //to functions taking double and
char  (*(*(* arr [3])(int))[4])(double) //returning char

:))

回答2:

Remember "delcaration mimics use". So to use said array you'd say

 (*FunctionPointers[0])();

Correct? Therefore to declare it, you use the same:

 void (*FunctionPointers[])() = { ... };

回答3:

Use this:

void (*FunctionPointers[])() = { };

Works like everything else, you place [] after the name.

来源：https://stackoverflow.com/questions/5093090/whats-the-syntax-for-declaring-an-array-of-function-pointers-without-using-a-se