对 C 限制限定符感到困惑

Question

首先，cppreference对restrict有如下说法：

During each execution of a block in which a restricted pointer P is declared (typically each execution of a function body in which P is a function parameter), if some object that is accessible through P (directly or indirectly) is modified, by any means, then all accesses to that object (both reads and writes) in that block must occur through P (directly or indirectly), otherwise the behavior is undefined.

但是在这一段下面写着：

Restricted pointers can be assigned to unrestricted pointers freely, the optimization opportunities remain in place as long as the compiler is able to analyze the code:

void f(int n, float * restrict r, float * restrict s) {
   float * p = r, * q = s; // OK
   while(n-- > 0) *p++ = *q++; // almost certainly optimized just like *r++ = *s++
}

在上面，r[0]是一个对象，可以通过受限指针r访问，它是通过p访问的，这似乎与第一段矛盾（那访问 r[0] 必须仅通过 r) ?

第二个：

Assignment from one restricted pointer to another is undefined behavior, except when assigning from a pointer to an object in some outer block to a pointer in some inner block (including using a restricted pointer argument when calling a function with a restricted pointer parameter) or when returning from a function (and otherwise when the block of the from-pointer ended).

那么，下面的代码可以吗？

void foo(char* restrict a, size_t s)
{
    for(char* restrict aa = a; aa < a + s; ++aa) // Is aa in an inner block?
    {
        *aa = '[=11=]';
    }
}

Answer 1

示例 1 的关键是用于访问 r 指向的底层数组的左值 *p 的地址基于 r。换句话说，它是通过 r 的间接访问。由于 *q 也有其基于 s 的地址，因此所有访问均通过原始受限指针发生，即使是间接访问：此处没有 UB。

示例 2 更为简单。您在原始指针上声明了一个新的受限指针 based。所以也没有UB。但是在示例 2 中，两个受限限定符都没有添加任何值，因为在块内部（foo 函数定义）只使用了一个指针，因为 aa 基于 a.

在示例 1 中，restrict 限定符向编译器声明由 r 和 s 指向的数组不重叠：无法通过 r 访问（直接或间接通过 p) 应该触及 s.

指向的数组

对 C 限制限定符感到困惑

Confused about C restrict qualifier

c

alias

restrict