在 C 和 C++ 中通过 index[array] 访问数组 [英] Accessing arrays by index[array] in C and C++

问题描述

有些面试官不管什么原因都喜欢问这个小技巧:

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?

据我所知，a[b] 被转换为 *(a + b) 并且由于加法是可交换的，所以它们的顺序并不重要，所以 2[a] 实际上是 *(2 + a) 并且效果很好.

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.

C 和/或 C++ 的规范保证这可以工作吗?

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

推荐答案

是的.6.5.2.1 第 1 段(C99 标准)描述了 [] 运算符的参数:

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

其中一个表达式的类型应为指向对象type的指针"，另一个表达式应为整数类型，结果的类型为type".

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 第 2 段(强调):

6.5.2.1 paragraph 2 (emphasis added):

后缀表达式后跟方括号中的表达式 [] 是下标数组对象的元素的指定.下标运算符[]的定义是 E1[E2] 与 (*((E1)+(E2))) 相同.由于转换规则适用于二进制 + 运算符，如果 E1 是一个数组对象(相当于，指向数组对象的初始元素)，E2 是一个整数，E1[E2] 指定 E2-thE1 的元素(从零开始计数).

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).

它没有说要求 [] 的参数顺序是合理的.

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

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