什么呢，行int *（*（X [3]）（））[5];用C呢？ [英] what does the line int *(*(x[3])())[5]; do in C?

问题描述

这是我第一次问上堆栈溢出的问题，所以请下跌自由地告诉我，如果我做错了什么，或不够具体。
我一直用C约4现在微控制器编程。几天前我在一家电子参加比赛。其中的许多问题是究竟是什么C $ C $克莱恩

This is my first time asking a question on Stack Overflow, so please fell free to tell me if I did anything wrong or not specific enough. I've been programming microcontrollers in C for about 4 now. Some days ago I took part in an Electronics Competition. One of the many questions was what exactly the C codeline

int *(*(x[3])())[5];

一样。这是该行，我记得。这可能是一个架在一个其他位置，但我认为这是行了。

does. This is the line as I remember it. It is possible that a bracket was at an other location, but I think this was the line.

我的猜测是，x是一个我们把第四个元素，并取消对它的引用functionpointers的数组。然后，这个函数时没有移交参数执行。的返回值似乎是一个指针的指针数组，其中我们解引用一次，以获得第一个元件的地址。然后，我们选择数组的元素第六。我不知道是什么int是因为虽然...

My guess is that x is an array of functionpointers of which we take the fourth element and dereference it. This function is then executed without handing over parameters. The returnvalue seems to be a pointer to a pointer an array of which we dereference once in order to get the address of the first element. We then choose the 6th element of that array. I have no idea what the int is for though...

非常感谢您回答我的问题，并有一个愉快的一天。

Thank you very much for answering my question and have a nice day.

推荐答案

阅读毛茸茸的声明，这样是出去找最左边的标识符和工作的方式，记住以下precedence规则的方式：

The way to read hairy declarations like this is to find the leftmost identifier and work your way out, remembering the following precedence rules:

*a[n]      -- a is an array of pointer
(*a)[n]    -- a is a pointer to an array
*f()       -- f is a function returning a pointer
(*f)()     -- f is a pointer to a function

应用这些规则，我们得到

Applying these rules, we get

        x             -- x is a
        x[3]          -- 3-element array of
       (x[3])()       -- function returning
      *(x[3])()       -- pointer to
     (*(x[3])())[5]   -- 5 element array of
    *(*(x[3])())[5]   -- pointer to
int *(*(x[3])())[5];  -- int

由于写的，这个声明是无效的;你不能有功能类型的阵列。我想这应该是像

As written, this declaration isn't valid; you can't have an array of function type. I imagine it was supposed to be something like

int *(*(*x[3])())[5];

其中， X 是一个数组的指针的久违指针的指针为int数组功能。

where x is an array of pointers to functions returning pointers to arrays of pointers to int.

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