这是如何与QUOT;阵列和QUOT的大小;模板函数的工作？ [英] How does this "size of array" template function work?
问题描述
可能显示的文件：结果
<一href=\"http://stackoverflow.com/questions/437150/can-someone-explain-this-template-$c$c-that-gives-me-the-size-of-an-array\">Can有人解释这个模板code，它给了我一个数组的大小？结果
魔术论点函数模板＆hellip;
有人能解释这是如何code的作品？我知道，这code的目的是为了获得一个数组的长度，但我不知道这是如何code工作：
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;}
感谢。
推荐答案
首先，让我们解剖参数， T（安培）【尺寸】。阅读声明从里面出来，从右到左，括号组第一：它的那就是大小尺寸类型 T的数组的引用一位不愿透露姓名的参数。
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.
即，它接受的任何阵列，其中该阵列的类型和尺寸是模板参数的参考。
That is, it accepts a reference to any array, where the type and size of the array are template parameters.
如果我们把它作为这样的：
If we call it as such:
int a[10];
GetArrLength(a);
编译器将尝试推断出模板参数。对于参数类型来匹配你传递什么， T 必须 INT 和尺寸必须是10（使参数数组引用10 INT 的）。
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 int's).
您再回到该大小，让您数组中元素的个数。
You then return that size, giving you the number of elements in an array.
有两个问题这个code。首先，大小不能为负，所以它没有意义使用符号的类型作为模板参数和返回类型。相反，一个无符号的类型应该使用;最好是的std ::为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，这个函数的结果是不恒定的前pression，即使一个数组大小。虽然这是在大多数情况下细，它会更好，如果我们能够从它那里得到一个恒定的前pression。这就是你结束了这个解决方案：
The second is that the result of this function is not a constant-expression, even though an arrays 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))
这被用作这样的：
int a[10];
const std::size_t n = sizeof_array(a); // constant-expression!
它的工作原理是三件事情：第一是同样的想法如上所示，模板参数将被填掉给你数组的大小
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.
第二部分是利用这些信息，使具有特定尺寸的类型，因此 type_of_size 帮手。这部分是不是绝对必要的，但我觉得它使code更容易阅读。 A 的char [N] 的大小等于 N ，一如既往，因此我们可以滥用，为存储的在一个类型本身！
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!
第三部分是让与这种规模的sizeof 。它实际上并没有什么评价，所以我们不需要为函数的定义。它只是说如果你做到这一点...规模将是...。而且规模是我们的存储的大小，字符阵列
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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