C函数指针语法

本文介绍了C函数指针语法的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

我的问题很简单。

通常，声明一些变量时，你把它的类型之前，如：

normally, when declaring some variable, you put its type before it, like:

int a;

一个函数指针可能拥有一个类型，如：INT（*）（INT，INT），在情况下，我们指出，它有两个int和返回一个int的函数。但是，声明这样的指针时，它的标识符不是类型后，如：

a function pointer may have type like: int(*)(int,int), in case we point to a function that takes two int and returns an int. But, when declaring such a pointer, its identifier is not after the type, like:

int(*)(int,int) mypointer;

相反，你必须写在中间的标识符：

instead, you must write the identifier in the middle:

int(*mypointer)(int,int);

为什么会这样？对不起，我知道这是一个embarassingly简单的问题...

why is this so? Sorry, I know it's an embarassingly easy question...

感谢大家的回答。A.S。

Thanks to everybody for replying. A.S.

推荐答案

我在回答Why是数组，指针，而这样设计的功能C语法，它基本上可以归结为：

I explain this in my answer to Why was the C syntax for arrays, pointers, and functions designed this way?, and it basically comes down to:

语言作者preferred使语法变量为中心，而不是类型为中心。也就是说，他们希望程序员看声明，并认为如果我写了前pression * FUNC（ARG），那将导致 INT ;如果我写 * ARG [N] 我将有一个浮动，而不是 FUNC 必须是指向一个功能采取的这个的和返回的是的。

the language authors preferred to make the syntax variable-centric rather than type-centric. That is, they wanted a programmer to look at the declaration and think "if I write the expression *func(arg), that'll result in an int; if I write *arg[N] I'll have a float" rather than "func must be a pointer to a function taking this and returning that".

这是维基百科的Ç进入声称：

里奇的想法是申报上下文识别类似的使用的声明反映使用

Ritchie's idea was to declare identifiers in contexts resembling their use: "declaration reflects use".

...援引K＆放P122; R2

...citing p122 of K&R2.