C++ 中指针运算的 a+i 和 &a[i] 有什么区别？

Question

假设我们有：

char* a;
int   i;

许多 C++ 介绍（如 this one) suggest that the rvalues a+i and &a[i] are interchangeable. I naively believed this for several decades, until I recently stumbled upon the following text (here) quoted from [dcl.ref]:

in particular, a null reference cannot exist in a well-defined program, because the only way to create such a reference would be to bind it to the "object" obtained by dereferencing a null pointer, which causes undefined behavior.

换句话说，"binding" 对空取消引用的引用对象会导致未定义的行为。基于 context of the above text, one infers that merely evaluating &a[i] (within the offsetof macro) is considered "binding" a reference. Furthermore, there seems to be a consensus that &a[i] causes undefined behavior in the case where a=null and i=0. This behavior is different from a+i (at least in C++, in the a=null, i=0 case).

这导致至少有 2 个关于 a+i 和 &a[i] 之间差异的问题：

首先，导致这种行为差异的 a+i 和 &a[i] 之间的潜在语义差异是什么。是否可以根据任何一种一般原则来解释，而不仅仅是 "binding a reference to a null dereference object causes undefined behavior just because this is a very specific case that everybody knows"？ &a[i] 是否会生成对 a[i] 的内存访问？或者规范作者那天对空引用不满意？或者别的什么？

其次，除了a=null和i=0的情况之外，还有其他情况a+i和&a[i]表现不同吗？（可以包含在第一个问题中，具体取决于对它的回答。）

Answer 1

在 C++ 标准中，[expr.sub]/1 部分您可以阅读：

The expression E1[E2] is identical (by definition) to *((E1)+(E2)).

这意味着 &a[i] 与 &*(a+i) 完全相同。因此，您首先要取消引用 * 一个指针，然后再获取地址 &。如果指针无效（即 nullptr，但也超出范围），则为 UB。

a+i 基于指针算法。起初它看起来不那么危险，因为没有取消引用肯定是 UB。不过也可能是UB（见[expr.add]/4:

When an expression that has integral type is added to or subtracted from a pointer, the result has the type of the pointer operand. If the expression P points to element x[i] of an array object x with n elements, the expressions P + J and J + P (where J has the value j) point to the (possibly-hypothetical) element x[i + j] if 0 ≤ i + j ≤ n; otherwise, the behavior is undefined. Likewise, the expression P - J points to the (possibly-hypothetical) element x[i − j] if 0 ≤ i − j ≤ n; otherwise, the behavior is undefined.

所以，虽然这两个表达式背后的语义略有不同，但我想说最终结果是一样的。

Answer 2

TL;DR：a+i 和 &a[i] 都是格式正确的，并且当 a 是空指针且 i 是 0 时产生空指针，根据符合标准（的意图），并且所有编译器都同意。

a+i 显然符合最新标准草案的 [expr.add]/4：

When an expression J that has integral type is added to or subtracted from an expression P of pointer type, the result has the type of P.

If P evaluates to a null pointer value and J evaluates to 0, the result is a null pointer value.

[...]

&a[i] 很棘手。 Per [expr.sub]/1, a[i] is equivalent to *(a+i), thus &a[i] is equivalent to &*(a+i). Now the standard is not quite clear about whether &*(a+i) is well-formed when a+i is a null pointer. But as @n.m. points out in , the intent as recorded in cwg 232 允许这种情况。

由于核心语言UB需要被常量表达式捕获([expr.const]/(4.6))，我们可以测试编译器是否认为这两个表达式是UB

这是演示，如果编译器认为 static_assert 中的常量表达式是 UB，或者如果他们认为结果不是 true，那么他们必须产生诊断（错误或警告）按照标准：

^{（请注意，这使用单参数 static_assert 和 constexpr lambda，它们是 C++17 的特性，默认的 lambda 参数也是很新的）}

static_assert(nullptr == [](char* a=nullptr, int i=0) {
    return a+i;
}());

static_assert(nullptr == [](char* a=nullptr, int i=0) {
    return &a[i];
}());

从 https://godbolt.org/z/hhsV4I 来看，似乎所有编译器在这种情况下都表现一致，根本不产生任何诊断（这让我有点惊讶）。

但是，这与 offset 的情况不同。 that question 中发布的实现明确创建了一个引用（这是回避用户定义 operator& 所必需的），因此受引用要求的约束。

C++ 中指针运算的 a+i 和 &a[i] 有什么区别？

What are the differences between a+i and &a[i] for pointer arithmetic in C++?

c++

language-lawyer

pointer-arithmetic