c - 为什么当 N 越界时 &a[N] 不调用 UB？

Question

我在很多例子中看到它是这样写的:

 #define N 5
 ....

 int a[N], *p;
 ....

 for (p = &a[0]; p < &a[N]; p++);

很明显 a[N] 不存在，那么为什么编译器不给出任何警告(如越界)或错误或者它调用 UB？

Answer 1

&a[N] 处的内存永远不会被您的程序访问，所以没问题。 C 标准允许比较数组对象内或数组对象末尾的指针。

编辑以下讨论:

&a[N] 不会导致未定义的行为 - 它完全等同于 a + N。来自 C 标准，6.5.3.2 地址和间接运算符，第 3 段:

The unary & operator yields the address of its operand … if the operand is the result of a [] operator, neither the & operator nor the unary * that is implied by the [] is evaluated and the result is as if the & operator were removed and the [] operator were changed to a + operator.