& a [n]是否有效，其中n是数组的大小? [英] Is &a[n] valid, where n is the size of the array?

问题描述

考虑以下简单代码对数组进行排序.

Consider the following simple code to sort an array.

int myarray[4] = {};
std::sort(myarray, myarray + 4);

我知道创建一个指向C样式数组末尾的指针是有效的.

I know that it is valid to create a pointer to one past the end of a C-style array.

我最近看过这样的代码:

I've recently seen code like this:

std::sort(myarray, &myarray[4]);

我不确定这是否有效，因为即使元素值未用于任何东西，它也会取消引用数组边界之外的元素.

I'm not sure this is valid, because it dereferences an element outside the array bounds, even though the element value is not used for anything.

这是有效的代码吗?

推荐答案

A[i]在语法上等效于*(A + i)，用于数组或指针A.
因此&A[i]在语法上等同于&(*(A + i)).

A[i] is syntactically equivalent to *(A + i) for an array or pointer A.
So &A[i] is syntactically equivalent to &(*(A + i)).

当*(A + i)没有未定义的行为时，&(*(A + i))的行为将与A + i相同.

When *(A + i) does not have undefined behavior, &(*(A + i)) will behave identically to A + i.

问题在于myarray[4]在语法上等同于*(myarray + 4)，后者会取消引用数组边界之外的位置.根据标准，这是未定义的行为.

The problem is that myarray[4] is syntactically equivalent to *(myarray + 4), which dereferences a location out of the array's bounds. That is undefined behavior according to the standard.

因此，您绝对应该优先选择myarray + 4而不是&myarray[4]-后者是未定义的行为.

So you should absolutely prefer myarray + 4 over &myarray[4] - the latter is undefined behavior.

&myarray[4]具有"正确"行为，大多数(如果不是全部的话)编译器并没有根据标准豁免其具有 undefined 行为.

That &myarray[4] has "correct" behavior with most - if not all - compilers does not exempt it from having undefined behavior according to the standard.

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