这个“数组大小"是如何计算的?模板功能有效吗? [英] How does this "size of array" template function work?

问题描述

有人可以解释一下这段代码是如何工作的吗?我知道这段代码的目的是获取数组的长度，但我不知道这段代码是如何工作的:

Can someone explain how this code works? I know that the purpose of this code is to get the length of an array, but I don't know how this code works:

template<typename T, int size>
int GetArrLength(T(&)[size]){return size;}

推荐答案

首先让我们剖析一下参数，T(&)[size].从内到外，从右到左，先读括号组的声明:这是一个未命名的参数，它是对 T 类型的大小为 size 的数组的引用.

First let's dissect the parameter, T(&)[size]. Read declarations from inside out, right to left, parenthesis group first: It's an unnamed parameter that is a reference to an array of size size of type T.

也就是说，它接受对任何数组的引用，其中数组的类型和大小是模板参数.

That is, it accepts a reference to any array, where the type and size of the array are template parameters.

如果我们这样称呼它:

int a[10];
GetArrLength(a);

编译器会尝试推断模板参数.要使参数类型与您传递的内容相匹配，T 必须是 int 并且 size 必须是 10(使参数成为对10 个 int 的数组).

The compiler will try to deduce the template parameters. For the parameter type to match what you're passing, T must be int and size must be 10 (making the parameter a reference to an array of 10 ints).

然后返回该大小，给出数组中元素的数量.

You then return that size, giving you the number of elements in an array.

这段代码有两个问题".首先，大小不能为负，因此使用带符号类型作为模板参数和返回类型是没有意义的.相反，应该使用无符号类型；最好是 std::size_t:

There are two "problems" with this code. Firstly, sizes cannot be negative, so it doesn't make sense to use a signed type as the template parameter and return type. Rather, an unsigned type should be used; best would be std::size_t:

template<typename T, std::size_t Size>
std::size_t GetArrLength(T(&)[Size]) { return size; }

第二个是这个函数的结果不是一个常量表达式，即使数组的大小是.虽然在大多数情况下这很好，但如果我们能从中得到一个常量表达式会更好.这就是您最终获得此解决方案的地方:

The second is that the result of this function is not a constant-expression, even though an array's size is. While that's fine in most situations, it would be better if we could get a constant-expression from it. That's where you end up with this solution:

template <std::size_t N>
struct type_of_size
{
    typedef char type[N];
};

template <typename T, std::size_t Size>
typename type_of_size<Size>::type& sizeof_array_helper(T(&)[Size]);

#define sizeof_array(pArray) sizeof(sizeof_array_helper(pArray))

这是这样使用的:

int a[10];
const std::size_t n = sizeof_array(a); // constant-expression!

它通过三件事起作用:第一件事与上面的想法相同，模板参数将被填写给你数组的大小.

It works by three things: the first is the same idea as above, that template parameters will be filled out giving you the array's size.

第二部分是使用该信息来创建具有特定大小的类型，因此是 type_of_size 助手.这部分不是绝对必要的，但我认为它使代码更易于阅读.char[N] 的大小总是等于 N，因此我们可以滥用它来存储"数组的大小......输入本身！

The second part is using that information to make a type with a specific size, hence the type_of_size helper. That part isn't strictly necessary, but I think it makes the code easier to read. A char[N] has a size equal to N, always, hence we can abuse that to "store" the size of the array...in the size of a type itself!

第三部分是使用 sizeof 获取该大小.它实际上不评估任何东西，所以我们不需要函数的定义.它只是说如果你这样做......大小会......".大小是我们存储"的大小，在 char 数组中.

The third part is getting that size with sizeof. It doesn't actually evaluate anything, so we don't need a definition for the function. It simply says "If you were to do this...the size would be...". And the size is our "stored" size, in the char array.

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