问题描述
假设我们有一个包含N个类型为T的元素的数组.
Assume we have an array that contains N elements of type T.
T a[N];
根据C ++ 14标准，在这种情况下，我们可以保证
According to the C++14 Standard, under which conditions do we have a guarantee that
(char*)(void*)&a[0] + n*sizeof(T) == (char*)(void*)&a[n], (0<=n<N) ?
尽管对于许多类型和实现都是如此，但该标准在脚注中并以一种含糊的方式提及了这一点:
While this is true for many types and implementations, the standard mentions it in a footnote, and in an ambiguous way:
几乎没有迹象表明其他方式被认为等同于标准方式.对于实施者而言，它可能暗示了许多一致的实施中的一种.
There is little indication that this other way was thought of being equivalent to the standard's way. It might rather be a hint for implementers that suggests one of many conforming implementations.
人们低估了这个问题的难度.
People have underestimated the difficulty of this question.
这个问题不是关于您在教科书中可以读到的内容，而是关于您可以通过使用逻辑和理由从C ++ 14标准中得出什么.
This question is not about what you can read in textbooks, it is about what what you can deduce from the C++14 Standard through the use of logic and reason.
如果您使用连续"或连续"，还请说出连续的内容.
If you use 'contiguous' or 'contiguously', please also say what is being contiguous.
虽然T []和T *密切相关，但是它们是抽象的，并且T * x N上的加法可以由实现以任何一致的方式定义.
While T[] and T* are closely related, they are abstractions, and the addition on T* x N may be defined by the implementation in any consistent way.
使用指针加法重新排列了方程式.如果p指向char，则始终使用(§5.7(4))或一元加法来定义p + 1，因此我们不会遇到UB.原始文件包含指针减法，这可能早早导致了UB. (只比较char指针，不取消引用).
The equation was rearranged using pointer addition. If p points to a char, p+1 is always defined using (§5.7 (4)) or unary addition, so we don't run into UB. The original included a pointer subtraction, which might have caused UB early on. (The char pointers are only compared, not dereferenced).
推荐答案
在[dcl.array]中:
In [dcl.array]:
连续表示T类型的任何连续子对象之间的偏移为sizeof(T)，这表示第n个子对象的偏移为n*sizeof(T).
Contiguous implies that the offset between any consecutive subobjects of type T is sizeof(T), which implies that the offset of the nth subobject is n*sizeof(T).
n < N的上限来自[expr.add]:
The upper bound of n < N comes from [expr.add]:
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