这一行声明了一个函数吗？ C++

发布于 2024-07-25 16:24:05 字数 524 浏览 4 评论 0原文

我正在阅读litb关于SFINAE的问题这里并且我想知道他的代码到底声明了什么。下面是一个更简单的（没有模板）示例：

int (&a())[2];

该声明到底是什么？ & 的作用是什么？更让我困惑的是，如果我声明以下内容，

int b()[2];

则会收到有关声明返回数组的函数的错误，而第一行没有此类错误（因此，人们会认为第一个声明不是函数）。但是，如果我尝试分配 a，

a = a;

则会收到一条错误消息，提示我正在尝试分配函数 a...，所以现在它是一个函数。这东西到底是什么？

原文

I was reading litb's question about SFINAE here and I was wondering exactly what his code is declaring. A simpler (without the templates) example is below:

int (&a())[2];

What exactly is that declaring? What is the role of the &? To add to my confusion, if I declare the following instead

int b()[2];

I get an error about declaring a function that returns an array, while the first line has no such error (therefore, one would think the first declaration is not a function). However, if I try to assign a

a = a;

I get an error saying I'm attempting to assign the function a... so now it is a function. What exactly is this thing?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

我一向站在原地 2024-08-01 16:24:05

有一些很棒的程序，称为 cdecl 和 c++decl。它们对于弄清楚复杂的声明非常有帮助，特别是对于 C 和 C++ 用于函数指针的拜占庭形式。

tyler@kusari ~ $ c++decl
Type `help' or `?' for help
c++decl> explain int (&a())[2]
declare a as function returning reference to array 2 of int
c++decl> explain int b()[2]
declare b as function returning array 2 of int

a 返回引用，b 不返回。

There's these awesome programs called cdecl and c++decl. They're very helpful for figuring out complicated declarations, especially for the byzantine forms that C and C++ use for function pointers.

tyler@kusari ~ $ c++decl
Type `help' or `?' for help
c++decl> explain int (&a())[2]
declare a as function returning reference to array 2 of int
c++decl> explain int b()[2]
declare b as function returning array 2 of int

a returns a reference, b does not.

回复收藏 0 原文

与之呼应 2024-08-01 16:24:05

为了便于将来参考，当您有一个特别难以解读的 C/C++ 声明时，您可能会发现此链接很有帮助：

如何阅读 C 声明

为了完整起见，我将重复其他人所说的内容来直接回答您的问题。

int (&a())[2];

...声明 a 为一个零参数函数，它返回对大小为 2 的整数数组的引用。（阅读上面链接上的基本规则，以清楚地了解我是如何想到的）

int b()[2];

...声明 b 是一个零参数函数，它返回一个大小为 2 的整数数组。

希望这可以帮助。

For future reference, you may find this link helpful when you have a particularly difficult C/C++ declaration to decipher:

How To Read C Declarations

For completeness, I will repeat what others have said to directly answer your question.

int (&a())[2];

...declares a to be a zero-argument function which returns a reference to an integer array of size 2. (Read the basic rules on the link above to have a clear understanding of how I came up with that.)

int b()[2];

...declares b to be a zero-argument function which returns an integer array of size two.

Hope this helps.

回复收藏 0 原文

弄潮 2024-08-01 16:24:05

int (&a())[2];

它声明了一个符号a，它是一个不带参数并返回对两元素整数数组的引用的函数。

 int b()[2];

这声明了一个符号b，它是一个不带参数并返回一个二元素整数数组的函数……这在语言的设计中是不可能的。

它相对简单：获取运算符优先级图表，以符号名称 (a) 开头，然后根据运算符的优先级开始应用运算符。每次应用操作后记下。

int (&a())[2];

It declares a symbol a that is a function that takes no arguments and returns a reference to a two-element array of integers.

 int b()[2];

This declares a symbol b that is a function that takes no arguments and returns a two-element array of integers... this is impossible by the design of the language.

It is relatively simple: get an operator precedence chart, start the symbol name (a) and start applying the operators as you see from their precedence. Write down after each operation applied.

回复收藏 0 原文

~没有更多了~