声明函数指针返回数组 [英] Declaring a function pointer returning an array

问题描述

为了练习，我正在尝试:

For practice, I'm trying to :

声明fubar为指向函数的指针，该函数采用指向char的指针并返回指向24个元素的数组的指针，其中每个元素均为指向struct foo的指针.

Declare fubar to be a pointer to a function that takes a pointer to a char and returns a pointer to an array of 24 elements where each element is a pointer to a struct foo.

我的逻辑是:

-fubar是指向带有char指针的函数的指针: (*fubar)(char*)

-fubar is a pointer to a function taking a char pointer: (*fubar)(char*)

-...返回一个指向24个元素的数组的指针，每个元素是一个结构foo:

-...returning a pointer to an array of 24 elems of where each elem is a struct foo:

(struct foo *)(*fubar)(char*)[24]

我的逻辑正确吗?

推荐答案

修复语法错误并删除struct foo *周围的括号后，

After fixing the syntax error and removing the parentheses around struct foo *,

struct foo* (*fubar)(char*)[24]

...您弄错的一部分是它实际上返回了一个 of 指针数组，而不是一个指向数组的指针.为了声明一个指向数组的指针，您需要额外的一组括号:

...the one part that you got wrong is that it actually returns an array of pointers, not a pointer to an array. In order to declare a pointer to the array, you need an extra set of parentheses:

struct foo (*(*fubar)(char*))[24]

您可以假装星星属于括号内的标识符(即数组的名称).

You can pretend that that the star belongs to an identifier (i.e., the name of the array) inside the parentheses.

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