pointers - 指向 Result, _> 中向量第一个元素的指针已损坏

Question

我有一个 Result<Vec<f64>, _> .当我尝试提取指向实际 f64 的指针时数组，我观察到的是数组 dptr指向的是预期数组的损坏版本(前 10 个字节已更改)。

为什么会发生这种情况，我该如何避免？

use std::error::Error;

fn main() {
    let res: Result<Vec<f64>, Box<dyn Error>> = Ok(vec![1., 2., 3., 4.]);
    let dptr: *const f64 = match res {
        Ok(v) => &v[0], 
        Err(_) => std::ptr::null(),
    };
    assert_eq!(unsafe { *dptr }, 1.0);
}

结果:

thread 'main' panicked at 'assertion failed: `(left == right)`
  left: `0.0`,
 right: `1.0`', src/main.rs:9:5

Playground

Answer 1

该程序的行为是未定义的，可以通过在 Miri 下运行它来查看，Miri 是一个 Rust 解释器，有时可以检测到未定义的行为。 (您可以在 playground 中通过单击“工具”(右上角)->“Miri”来执行此操作):

error: Undefined Behavior: pointer to alloc1039 was dereferenced after this allocation got freed
 --> src/main.rs:9:25
  |
9 |     assert_eq!(unsafe { *dptr }, 1.0);
  |                         ^^^^^ pointer to alloc1039 was dereferenced after this allocation got freed
  |
  = help: this indicates a bug in the program: it performed an invalid operation, and caused Undefined Behavior
  = help: see https://doc.rust-lang.org/nightly/reference/behavior-considered-undefined.html for further information

这里发生的是释放后使用:Ok(v) => &v[0], 行从 v 移动数据(因此 res)，导致它被释放。后来，用于其他变量的数据覆盖了现有数据(因为在它指向的内存被释放后使用指针是未定义的行为)。

如果您尝试在没有 unsafe 的情况下以正常方式从 res 中读取数据，那么对于这个问题您会遇到编译时错误:

error[E0382]: use of partially moved value: `res`
 --> src/main.rs:9:10
  |
6 |         Ok(v) => &v[0],
  |            - value partially moved here
...
9 |     dbg!(res.unwrap()[0]);
  |          ^^^ value used here after partial move
  |
  = note: partial move occurs because value has type `Vec<f64>`, which does not implement the `Copy` trait
help: borrow this field in the pattern to avoid moving `res.0`
  |
6 |         Ok(ref v) => &v[0],
  |            +++

(playground)