C++ 在不使用单独的 typedef 的情况下声明函数指针数组的语法是什么？
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What's the syntax for declaring an array of function pointers without using a separate typedef?
提问by Maxpm
Arrays of function pointers can be created like so:
可以像这样创建函数指针数组：
typedef void(*FunctionPointer)();
FunctionPointer functionPointers[] = {/* Stuff here */};
What is the syntax for creating a function pointer array without using the typedef?
在不使用typedef? 的情况下创建函数指针数组的语法是什么？
回答by Armen Tsirunyan
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
所以你的答案是
void (* arr [])() = {};
But naturally, this is a bad practice, just use typedefs:)
但自然，这是一个不好的做法，只需使用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? :))
额外：想知道如何声明一个包含 3 个指针的数组，该数组指向采用 int 的函数并返回一个指向 4 个指向采用 double 并返回 char 的函数的指针数组的指针？（这有多酷，嗯？:)）
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by Logan Capaldo
Remember "delcaration mimics use". So to use said array you'd say
记住“声明模仿使用”。所以要使用所说的数组，你会说
(*FunctionPointers[0])();
Correct? Therefore to declare it, you use the same:
正确的？因此要声明它，您使用相同的：
void (*FunctionPointers[])() = { ... };
回答by Erik
Use this:
用这个：
void (*FunctionPointers[])() = { };
Works like everything else, you place [] after the name.
像其他一切一样工作，您将 [] 放在名称后面。
