std::end怎么知道数组的结束?-Howdoesstd::endknowtheendofanarray?

22  

is the constant integer 5 will be stored some where?

常数整数5会被存储在什么地方?

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

是的，它是数组类型的一部分。但不，它没有显式地存储在任何地方。当你有

int i[5] = { };

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

i的类型是int[5]。沙菲克的回答谈到了如何使用这个长度来实现end。

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

如果您有c++ 11，那么使用constexpr将是最简单的方法

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

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:

如果您有一个前c++ 11编译器，其中constexpr不可用，那么上面的函数可能不会在编译时进行计算。所以在这种情况下，你可以用这个:

template 
char (&arrLenFn(const T (&arr) [N]))[N];

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

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.

首先，我们声明一个函数返回一个对N个字符的数组的引用，例如，这个函数的sizeof现在是数组的长度。然后我们有一个宏来包装它，这样它在调用者的端是可读的。

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.

注意:两个基类型相同但长度不同的数组仍然是两个完全不同的类型。int[3]不同于int[2]。然而，在这两种情况下，数组衰变都会使您得到int*。阅读如何在c++中使用数组?如果你想知道更多。

25  

But, how does it know the end of an array

但是，它如何知道数组的结束

It uses a template non-type parameter to deduce the size of the array, which can then be used to produce the end pointer. The C++11 signature from the cppreference section for std::end is as follows:

它使用一个模板非类型参数来推断数组的大小，然后可以使用它来生成最终指针。std:::的cppreference部分的c++ 11签名如下:

template
T* end( T (&array)[N] );

As hvd notes, since it is passed by reference this prevents decay to a pointer.

正如hvd所指出的，由于它是通过引用传递的，因此可以防止对指针的衰减。

The implementation would be something similar to:

执行情况将类似于:

template
T* end( T (&array)[N] )
{
    return array + N ;
}

Is the constant integer 5 will be stored some where?

常数整数5会被存储在什么地方?

5 or N is part of the type of the array and so N is available at compile time. For example applying sizeof to an array will give us the total number of bytes in the array.

5或N是数组类型的一部分，所以N在编译时可用。例如，将sizeof应用到一个数组中将给出数组中字节的总数。

Many times we see an array passed by value to a function. In that case, the array decays to a pointer to type stored in the array. So now the size information is lost. Passing by reference allows us to avoid this loss of information and extract the size N from the type.

很多时候我们看到一个数组通过值传递给一个函数。在这种情况下，数组会衰减到存储在数组中的类型的指针。所以现在大小信息丢失了。通过引用传递可以避免信息丢失，并从类型中提取大小为N的值。

8  

Because you are passing an array to std::end, and an array has type T [N]. std::end can tell when the array ends by looking at the N in the type.

因为您正在将一个数组传递给std::end，而一个数组的类型是T [N]。end可以通过查看类型中的N来判断数组何时结束。