Rust 将指向结构体第一个成员的指针转换为指向该结构体的指针是否合法？

Question

在 C 中，指向结构的指针可以转换为指向其第一个成员的指针，反之亦然。也就是说，结构体的地址被定义为其第一个成员的地址。

struct Base { int x; };
struct Derived { struct Base base; int y; };

int main() {
    struct Derived d = { {5}, 10 };
    struct Base *base = &d.base; // OK
    printf("%d\n", base->x);
    struct Derived *derived = (struct Derived *)base; // OK
    printf("%d\n", derived->y);
}

这通常用于实现 C++ 风格的继承。

如果结构是repr(C)（这样它们的字段就不会被重新组织），Rust 中是否允许同样的事情？

struct Base { int x; };
struct Derived { struct Base base; int y; };

int main() {
    struct Derived d = { {5}, 10 };
    struct Base *base = &d.base; // OK
    printf("%d\n", base->x);
    struct Derived *derived = (struct Derived *)base; // OK
    printf("%d\n", derived->y);
}

代码可以工作，但它会一直工作吗？它是定义的行为吗？

Answer 1

你写的方式，目前不是；但让它发挥作用是可能的。

对 的引用T仅允许访问T- 不再允许（它具有 的出处T）。该表达式为您提供了仅对 有效的&d.base参考。使用它来访问 的字段是未定义的行为。目前还不清楚这是否是我们想要的，并且对此有积极的讨论（还有this），但这就是当前的行为。有一个名为Miri的好工具，它允许您检查 Rust 代码是否存在某些（不是全部！）未定义的行为（您可以在 Playground 中运行它；工具 -> Miri），而且它确实会标记您的代码：BaseDerived

error: Undefined Behavior: trying to reborrow <untagged> for SharedReadOnly permission at alloc1707[0x8], but that tag does not exist in the borrow stack for this location
  --> src/main.rs:17:5
   |
17 |     &*ptr
   |     ^^^^^
   |     |
   |     trying to reborrow <untagged> for SharedReadOnly permission at alloc1707[0x8], but that tag does not exist in the borrow stack for this location
   |     this error occurs as part of a reborrow at alloc1707[0x0..0x10]
   |
   = help: this indicates a potential bug in the program: it performed an invalid operation, but the rules it violated are still experimental
   = help: see https://github.com/rust-lang/unsafe-code-guidelines/blob/master/wip/stacked-borrows.md for further information
           
   = note: inside `get_derived_from_base` at src/main.rs:17:5
note: inside `main` at src/main.rs:31:28
  --> src/main.rs:31:28
   |
31 |     let derived = unsafe { get_derived_from_base(base) }; // defined behaviour?
   |                            ^^^^^^^^^^^^^^^^^^^^^^^^^^^

Derived您可以通过创建对整体的引用并将其转换为指向 的原始指针来使其工作Base。原始指针将保留原始引用的出处，因此这将起作用：

error: Undefined Behavior: trying to reborrow <untagged> for SharedReadOnly permission at alloc1707[0x8], but that tag does not exist in the borrow stack for this location
  --> src/main.rs:17:5
   |
17 |     &*ptr
   |     ^^^^^
   |     |
   |     trying to reborrow <untagged> for SharedReadOnly permission at alloc1707[0x8], but that tag does not exist in the borrow stack for this location
   |     this error occurs as part of a reborrow at alloc1707[0x0..0x10]
   |
   = help: this indicates a potential bug in the program: it performed an invalid operation, but the rules it violated are still experimental
   = help: see https://github.com/rust-lang/unsafe-code-guidelines/blob/master/wip/stacked-borrows.md for further information
           
   = note: inside `get_derived_from_base` at src/main.rs:17:5
note: inside `main` at src/main.rs:31:28
  --> src/main.rs:31:28
   |
31 |     let derived = unsafe { get_derived_from_base(base) }; // defined behaviour?
   |                            ^^^^^^^^^^^^^^^^^^^^^^^^^^^

注意：在此过程中不应涉及参考文献。如果您重新借用base作为 type 的引用Base，您将再次失去来源。这将在操场上通过 Miri，但根据当前规则仍然是未定义的行为，并且会因更严格的标志而使 Miri 失败（在本地运行 Miri 之前将环境变量设置为MIRIFLAGS）-Zmiri-tag-raw-pointers。