std::end 如何知道数组的结尾? [英] How does std::end know the end of an array?

问题描述

std::begin 和 std::end 知道 container 或 array 的开始和结束>.

std::begin and std::end know the beginning and end of a container or an array.

例如，很容易知道 vector 的 end 和 begin，因为它是一个提供此信息的类.但是，它如何知道像下面这样的 array 的结尾?

It so easy to know the end and begin of a vector for example because it is a class that gives this information. But, how does it know the end of an array like the following?

int simple_array[5]{1, 2, 3, 4, 5};
auto beg=std::begin(simple_array);
auto en=std::end(simple_array);

std::begin 并不难知道数组从哪里开始.但是它怎么知道它在哪里结束呢?常量整数 5 会存储在某个地方吗?

std::begin is not that hard to know where the array start. But how does it know where it ends? Will the constant integer 5 be stored somewhere?

如果我能得到一些低级信息的答案，我将不胜感激.

I would appreciate if I got an answer with some low-level information.

推荐答案

常量整数 5 会存储在什么地方吗?

is the constant integer 5 will be stored some where?

是的，它是数组类型的一部分.但是不，它没有明确存储在任何地方.当你有

Yes, it's part of the type of the array. But no, it's not stored anywhere explicitly. When you have

int i[5] = { };

i 的类型是 int[5].Shafik 的回答谈到了如何使用这个长度来实现 end.

the type of i is int[5]. Shafik's answer talks about how this length is used to implement end.

如果你有 C++11，使用 constexpr 将是最简单的方法

If you've C++11, using constexpr would be the simple way to go

template <typename T, size_t N>
inline constexpr size_t
arrLen(const T (&arr) [N]) {
    return N;
}

如果您有一个 C++11 之前的编译器，其中 constexpr 不可用，则可能不会在编译时评估上述函数.所以在这种情况下，你可以使用这个:

If you've a pre-C++11 compiler where constexpr isn't available, the above function may not be evaluated at compile-time. So in such situations, you may use this:

template <typename T, size_t N>
char (&arrLenFn(const T (&arr) [N]))[N];

#define arrLen(arr) sizeof(arrLenFn(arr))

首先我们声明一个函数，返回对 N char 数组的引用，即 sizeof 这个函数现在是数组的长度.然后我们有一个宏来包装它，这样它在调用者端是可读的.

First we declare a function returning a reference to an array of N chars i.e. sizeof this function would now be the length of the array. Then we've a macro to wrap it, so that it's readable at the caller's end.

注意:两个基类型相同但长度不同的数组仍然是两种完全不同的类型.int[3] 与 int[2] 不同.数组衰减，然而，在这两种情况下都会得到 int*.如果您想了解更多信息，请阅读如何在 C++ 中使用数组?.

Note: Two arrays of the same base type but with different lengths are still two completely different types. int[3] is not the same as int[2]. Array decay, however, would get you an int* in both cases. Read How do I use arrays in C++? if you want to know more.

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