在C ++中是int [pointer-to-array]-标准吗? [英] is int[pointer-to-array] in the C++ - standard?

问题描述

据我所知，可以编写以下代码:

As I have learned, one can write the following code:

char *a = new char[50];
for (int i = 0; i < 50; ++i) {
    i[a] = '5';
}

它会编译.有用.与

完全相同

It compiles. It works. It does exactly the same as

char *a = new char[50];
for (int i = 0; i < 50; ++i) {
    a[i] = '5';
}

是因为:

默认情况下，
• a[b]被实现为宏*(a + b)，并且两个代码示例均有效，这只是偶然/编译器特有的
• 它在某处是标准化的，并且在每个平台上，此类算法的结果应该相同

• a[b] is implemented as a macro *(a + b) by default and the fact that both code samples are valid is just an accident/compiler specific
• it's standardized somewhere and the outcome of such algorithms should be the same on every platform

合理地假设加法应该是可交换的，但是如果我们以这种方式实现operator[]，则我们已经将其他东西进行了交换，这可能不是我们想要的.

It is reasonable to assume that addition should be commutative, but if we implement operator[] in that way, we have made something else commutative, what might not be what we wanted.

有趣的事实是，没有pointer[pointer]运算符，所以operator[]不是宏.

The interesting fact is that there is no pointer[pointer] operator, so operator[] is not a macro.

我知道这很糟糕.我知道这会使阅读代码的人感到困惑.但是我想知道这是否只是偶然的情况，并且在独角兽有七只脚和角在左脸的遥远土地上行不通.

I know it's bad. I know it's confusing the people who read the code. But I want to know if it's just an accident and it will not work in a distant land where unicorns have seven legs and horns are on their left cheek.

推荐答案

C ++标准，第8.3.4节，注释7(第185页)(重点是我的).

除已为类(13.5.5)声明的地方外，下标运算符[]的解释方式为E1[E2]与*((E1)+(E2))相同.由于适用于+的转换规则，如果E1是数组，而E2是整数，则E1[E2]引用E1的第E2个成员.因此，尽管外观不对称，但下标是可交换的操作.

Except where it has been declared for a class (13.5.5), the subscript operator [] is interpreted in such a way that E1[E2] is identical to *((E1)+(E2)). Because of the conversion rules that apply to +, if E1 is an array and E2 an integer, then E1[E2] refers to the E2-th member of E1. Therefore, despite its asymmetric appearance, subscripting is a commutative operation.

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