在C和C指数[数组]数组访问++ [英] Accessing arrays by index[array] in C and C++

问题描述

因此​​，有这个小动作问题，有些面试官喜欢问无论出于何种原因：

So, there is this little trick question that some interviewers like to ask for whatever reason:

int arr[] = {1, 2, 3};
2[arr] = 5; // does this line compile?
assert(arr[2] == 5); // does this assertion fail?

从我能理解， A [B] 被转换为 *（A + B）和自加法是可交换它并没有真正不管他们的顺序，因此 2 [A] 真的是 *（2 + A）和工作正常。

From what I can understand, a[b] gets converted to *(a + b) and since addition is commutative it doesn't really matter their order, so 2[a] is really *(2 + a) and that works fine.

这是通过保证C和/或C ++语言的规范工作？

Is this guaranteed to work by C and/or C++'s specs?

推荐答案

是的。 6.5.2.1款（C99标准）描述的参数在 [] 运营商：

Yes. 6.5.2.1 paragraph 1 (C99 standard) describes the arguments to the [] operator:

其中前pressions应该是一个类型的指针对象键入时，其他前pression应具有整数类型，结果有键入键入。

One of the expressions shall have type "pointer to object type", the other expression shall have integer type, and the result has type "type".

6.5.2.1条第2款（强调）：

6.5.2.1 paragraph 2 (emphasis added):

一个后连接点¯x前pression随后在方括号中前pression [] 是一个下标
  阵列对象的元素的指定。 下标运算符的定义 []
  是 E1 [E2] 与（*（（E1）+（E2））） 。由于转换规则的那
  适用于二进制 + 运营商，如果 E1 是一个数组对象（等同于一个指针
  数组对象的初始元素）和 E2 是一个整数， E1 [E2] 指定的 E2 -th
   E1 （从零算起）。元素

A postfix expression followed by an expression in square brackets [] is a subscripted designation of an element of an array object. The definition of the subscript operator [] is that E1[E2] is identical to (*((E1)+(E2))). Because of the conversion rules that apply to the binary + operator, if E1 is an array object (equivalently, a pointer to the initial element of an array object) and E2 is an integer, E1[E2] designates the E2-th element of E1 (counting from zero).

报告说，没有什么要求的参数以 [] 是明智的。

It says nothing requiring the order of the arguments to [] to be sane.

这篇关于在C和C指数[数组]数组访问++的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！