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<main>
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<h1 id="测试栈借用"><a class="header" href="#测试栈借用">测试栈借用</a></h1>
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<blockquote>
|
|
|
<p>关于上一章节的简单总结:</p>
|
|
|
<ul>
|
|
|
<li>Rust 通过借用栈来处理再借用</li>
|
|
|
<li>只有栈顶的元素是处于 <code>live</code> 状态的( 被借用 )</li>
|
|
|
<li>当访问栈顶下面的元素时，该元素会变为 <code>live</code>，而栈顶元素会被弹出( <code>pop</code> )</li>
|
|
|
<li>从借用栈中弹出的元素无法再被借用</li>
|
|
|
<li>借用检查器会保证我们的安全代码遵守以上规则</li>
|
|
|
<li>Miri 可以在一定程度上保证裸指针在运行时也遵循以上规则</li>
|
|
|
</ul>
|
|
|
</blockquote>
|
|
|
<p>作为作者同时也是读者，我想说上一章节的内容相当不好理解，下面来看一些例子，通过它们可以帮助大家更好的理解栈借用模型。</p>
|
|
|
<p>在实际项目中捕获 UB 是一件相当不容易的事，毕竟你是在编译器的盲区之外摸索和行动。</p>
|
|
|
<p>如果我们足够幸运的话，写出来的代码是可以"正常运行的“，但是一旦编译器聪明一点或者你修改了某处代码，那这些代码可能会立刻化身为一颗安静的定时炸弹。当然，如果你还是足够幸运，那程序会发生崩溃，你也就可以捕获和处理相应的错误。但是如果你不幸运呢？</p>
|
|
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<p>那代码就算出问题了，也只是会发生一些奇怪的现象，面对这些现象你将束手无策，甚至不知道该如何处理！</p>
|
|
|
<p>Miri 为何可以一定程度上提前发现这些 UB 问题？因为它会去获取 rustc 对我们的程序最原生、且没有任何优化的视角，然后对看到的内容进行解释和跟踪。只要这个过程能够开始，那这个解决方法就相当有效，但是问题来了，该如何让这个过程开始？要知道 Miri 和 rustc 是不可能去逐行分析代码中的所有行为的，这样做的结果就是编译时间大大增加！</p>
|
|
|
<p>因此我们需要使用测试用例来让程序中可能包含 UB 的代码路径被真正执行到，当然，就算你这么做了，也不能完全依赖 Miri。既然是分析，就有可能遗漏，也可能误杀友军。</p>
|
|
|
<h2 id="基本借用"><a class="header" href="#基本借用">基本借用</a></h2>
|
|
|
<p>在上一章节中，借用检查器似乎不喜欢以下代码:</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>let mut data = 10;
|
|
|
let ref1 = &mut data;
|
|
|
let ref2 = &mut *ref1;
|
|
|
|
|
|
*ref1 += 1;
|
|
|
*ref2 += 2;
|
|
|
|
|
|
println!("{}", data);
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<p>它违背了再借用的原则，大家可以用借用栈的分析方式去验证下上一章节所学的知识。</p>
|
|
|
<p>下面来看看，如果使用裸指针会怎么样:</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>unsafe {
|
|
|
let mut data = 10;
|
|
|
let ref1 = &mut data;
|
|
|
let ptr2 = ref1 as *mut _;
|
|
|
|
|
|
*ref1 += 1;
|
|
|
*ptr2 += 2;
|
|
|
|
|
|
println!("{}", data);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
Compiling miri-sandbox v0.1.0
|
|
|
Finished dev [unoptimized + debuginfo] target(s) in 0.71s
|
|
|
Running `target\debug\miri-sandbox.exe`
|
|
|
13
|
|
|
</code></pre>
|
|
|
<p>嗯，编译器看起来很满意：不仅获取了预期的结果，还没有任何警告。那么再来征求下 Miri 的意见：</p>
|
|
|
<pre><code class="language-shell">MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
|
|
|
Finished dev [unoptimized + debuginfo] target(s) in 0.00s
|
|
|
Running cargo-miri.exe target\miri
|
|
|
|
|
|
error: Undefined Behavior: no item granting read access
|
|
|
to tag <untagged> at alloc748 found in borrow stack.
|
|
|
|
|
|
--> src\main.rs:9:9
|
|
|
|
|
|
|
9 | *ptr2 += 2;
|
|
|
| ^^^^^^^^^^ no item granting read access to tag <untagged>
|
|
|
| at alloc748 found in borrow stack.
|
|
|
|
|
|
|
= help: this indicates a potential bug in the program:
|
|
|
it performed an invalid operation, but the rules it
|
|
|
violated are still experimental
|
|
|
</code></pre>
|
|
|
<p>喔，果然出问题了。下面再来试试更复杂的 <code>&mut -> *mut -> &mut -> *mut</code> :</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>unsafe {
|
|
|
let mut data = 10;
|
|
|
let ref1 = &mut data;
|
|
|
let ptr2 = ref1 as *mut _;
|
|
|
let ref3 = &mut *ptr2;
|
|
|
let ptr4 = ref3 as *mut _;
|
|
|
|
|
|
// 首先访问第一个裸指针
|
|
|
*ptr2 += 2;
|
|
|
|
|
|
// 接着按照借用栈的顺序来访问
|
|
|
*ptr4 += 4;
|
|
|
*ref3 += 3;
|
|
|
*ptr2 += 2;
|
|
|
*ref1 += 1;
|
|
|
|
|
|
println!("{}", data);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
22
|
|
|
</code></pre>
|
|
|
<pre><code class="language-shell">MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
|
|
|
error: Undefined Behavior: no item granting read access
|
|
|
to tag <1621> at alloc748 found in borrow stack.
|
|
|
|
|
|
--> src\main.rs:13:5
|
|
|
|
|
|
|
13 | *ptr4 += 4;
|
|
|
| ^^^^^^^^^^ no item granting read access to tag <1621>
|
|
|
| at alloc748 found in borrow stack.
|
|
|
|
|
|
|
</code></pre>
|
|
|
<p>不错，可以看出 miri 有能力分辨两个裸指针的使用限制：当使用第二个时，需要先让之前的失效。</p>
|
|
|
<p>再来移除乱入的那一行，让借用栈可以真正顺利的工作：</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>unsafe {
|
|
|
let mut data = 10;
|
|
|
let ref1 = &mut data;
|
|
|
let ptr2 = ref1 as *mut _;
|
|
|
let ref3 = &mut *ptr2;
|
|
|
let ptr4 = ref3 as *mut _;
|
|
|
|
|
|
// Access things in "borrow stack" order
|
|
|
*ptr4 += 4;
|
|
|
*ref3 += 3;
|
|
|
*ptr2 += 2;
|
|
|
*ref1 += 1;
|
|
|
|
|
|
println!("{}", data);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
20
|
|
|
|
|
|
MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
20
|
|
|
</code></pre>
|
|
|
<p>我现在可以负责任的说：在座的各位，都是...可以获取编程语言内存模型设计博士学位的存在，编译器？那是什么东东！简单的很。</p>
|
|
|
<blockquote>
|
|
|
<p>旁白：那个..关于博士的一切，请不要当真，但是我依然为你们骄傲</p>
|
|
|
</blockquote>
|
|
|
<h2 id="测试数组"><a class="header" href="#测试数组">测试数组</a></h2>
|
|
|
<p>下面来干一票大的：使用指针偏移来搞乱一个数组。</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>unsafe {
|
|
|
let mut data = [0; 10];
|
|
|
let ref1_at_0 = &mut data[0]; // 获取第 1 个元素的引用
|
|
|
let ptr2_at_0 = ref1_at_0 as *mut i32; // 裸指针 ptr 指向第 1 个元素
|
|
|
let ptr3_at_1 = ptr2_at_0.add(1); // 对裸指针进行运算，指向第 2 个元素
|
|
|
|
|
|
*ptr3_at_1 += 3;
|
|
|
*ptr2_at_0 += 2;
|
|
|
*ref1_at_0 += 1;
|
|
|
|
|
|
// Should be [3, 3, 0, ...]
|
|
|
println!("{:?}", &data[..]);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
[3, 3, 0, 0, 0, 0, 0, 0, 0, 0]
|
|
|
</code></pre>
|
|
|
<pre><code class="language-shell">MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
|
|
|
error: Undefined Behavior: no item granting read access
|
|
|
to tag <1619> at alloc748+0x4 found in borrow stack.
|
|
|
--> src\main.rs:8:5
|
|
|
|
|
|
|
8 | *ptr3_at_1 += 3;
|
|
|
| ^^^^^^^^^^^^^^^ no item granting read access to tag <1619>
|
|
|
| at alloc748+0x4 found in borrow stack.
|
|
|
</code></pre>
|
|
|
<p>咦？我们命名按照借用栈的方式来完美使用了，为何 miri 还是提示了 UB 风险？难道是因为 <code>ptr -> ptr</code> 的过程中发生了什么奇怪的事情？如果我们只是拷贝指针，让它们都指向同一个位置呢？</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>unsafe {
|
|
|
let mut data = [0; 10];
|
|
|
let ref1_at_0 = &mut data[0];
|
|
|
let ptr2_at_0 = ref1_at_0 as *mut i32;
|
|
|
let ptr3_at_0 = ptr2_at_0;
|
|
|
|
|
|
*ptr3_at_0 += 3;
|
|
|
*ptr2_at_0 += 2;
|
|
|
*ref1_at_0 += 1;
|
|
|
|
|
|
// Should be [6, 0, 0, ...]
|
|
|
println!("{:?}", &data[..]);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
[6, 0, 0, 0, 0, 0, 0, 0, 0, 0]
|
|
|
|
|
|
MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
[6, 0, 0, 0, 0, 0, 0, 0, 0, 0]
|
|
|
</code></pre>
|
|
|
<p>果然，顺利通过，下面我们还是让它们指向同一个位置，但是来首名为混乱的 BGM:</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>unsafe {
|
|
|
let mut data = [0; 10];
|
|
|
let ref1_at_0 = &mut data[0]; // Reference to 0th element
|
|
|
let ptr2_at_0 = ref1_at_0 as *mut i32; // Ptr to 0th element
|
|
|
let ptr3_at_0 = ptr2_at_0; // Ptr to 0th element
|
|
|
let ptr4_at_0 = ptr2_at_0.add(0); // Ptr to 0th element
|
|
|
let ptr5_at_0 = ptr3_at_0.add(1).sub(1); // Ptr to 0th element
|
|
|
|
|
|
|
|
|
*ptr3_at_0 += 3;
|
|
|
*ptr2_at_0 += 2;
|
|
|
*ptr4_at_0 += 4;
|
|
|
*ptr5_at_0 += 5;
|
|
|
*ptr3_at_0 += 3;
|
|
|
*ptr2_at_0 += 2;
|
|
|
*ref1_at_0 += 1;
|
|
|
|
|
|
// Should be [20, 0, 0, ...]
|
|
|
println!("{:?}", &data[..]);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
[20, 0, 0, 0, 0, 0, 0, 0, 0, 0]
|
|
|
|
|
|
MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
[20, 0, 0, 0, 0, 0, 0, 0, 0, 0]
|
|
|
</code></pre>
|
|
|
<p>可以看出，<code>miri</code> 对于这种裸指针派生是相当纵容的：当它们都共享同一个借用时(borrowing, 也可以用 miri 的称呼： tag)。</p>
|
|
|
<blockquote>
|
|
|
<p>当代码足够简单时，编译器是有可能介入跟踪所有派生的裸指针，并尽可能去优化它们的。但是这套规则比引用的那套脆弱得多！</p>
|
|
|
</blockquote>
|
|
|
<p>那么问题来了：真正的问题到底是什么？</p>
|
|
|
<p>对于部分数据结构，Rust 允许对其中的字段进行独立借用，例如一个结构体，它的多个字段可以被分开借用，来试试这里的数组可不可以。</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>unsafe {
|
|
|
let mut data = [0; 10];
|
|
|
let ref1_at_0 = &mut data[0]; // Reference to 0th element
|
|
|
let ref2_at_1 = &mut data[1]; // Reference to 1th element
|
|
|
let ptr3_at_0 = ref1_at_0 as *mut i32; // Ptr to 0th element
|
|
|
let ptr4_at_1 = ref2_at_1 as *mut i32; // Ptr to 1th element
|
|
|
|
|
|
*ptr4_at_1 += 4;
|
|
|
*ptr3_at_0 += 3;
|
|
|
*ref2_at_1 += 2;
|
|
|
*ref1_at_0 += 1;
|
|
|
|
|
|
// Should be [3, 3, 0, ...]
|
|
|
println!("{:?}", &data[..]);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">error[E0499]: cannot borrow `data[_]` as mutable more than once at a time
|
|
|
--> src\main.rs:5:21
|
|
|
|
|
|
|
4 | let ref1_at_0 = &mut data[0]; // Reference to 0th element
|
|
|
| ------------ first mutable borrow occurs here
|
|
|
5 | let ref2_at_1 = &mut data[1]; // Reference to 1th element
|
|
|
| ^^^^^^^^^^^^ second mutable borrow occurs here
|
|
|
6 | let ptr3_at_0 = ref1_at_0 as *mut i32; // Ptr to 0th element
|
|
|
| --------- first borrow later used here
|
|
|
|
|
|
|
= help: consider using `.split_at_mut(position)` or similar method
|
|
|
to obtain two mutable non-overlapping sub-slices
|
|
|
</code></pre>
|
|
|
<p>显然..不行，Rust 不允许我们对数组的不同元素进行单独的借用，注意到提示了吗？可以使用 <code>.split_at_mut(position)</code> 来将一个数组分成多个部分:</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>unsafe {
|
|
|
let mut data = [0; 10];
|
|
|
|
|
|
let slice1 = &mut data[..];
|
|
|
let (slice2_at_0, slice3_at_1) = slice1.split_at_mut(1);
|
|
|
|
|
|
let ref4_at_0 = &mut slice2_at_0[0]; // Reference to 0th element
|
|
|
let ref5_at_1 = &mut slice3_at_1[0]; // Reference to 1th element
|
|
|
let ptr6_at_0 = ref4_at_0 as *mut i32; // Ptr to 0th element
|
|
|
let ptr7_at_1 = ref5_at_1 as *mut i32; // Ptr to 1th element
|
|
|
|
|
|
*ptr7_at_1 += 7;
|
|
|
*ptr6_at_0 += 6;
|
|
|
*ref5_at_1 += 5;
|
|
|
*ref4_at_0 += 4;
|
|
|
|
|
|
// Should be [10, 12, 0, ...]
|
|
|
println!("{:?}", &data[..]);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
[10, 12, 0, 0, 0, 0, 0, 0, 0, 0]
|
|
|
|
|
|
MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
[10, 12, 0, 0, 0, 0, 0, 0, 0, 0]
|
|
|
</code></pre>
|
|
|
<p>将数组切分成两个部分后，代码就成功了，如果我们将一个切片转换成指针呢？那指针是否还拥有访问整个切片的权限？</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>unsafe {
|
|
|
let mut data = [0; 10];
|
|
|
|
|
|
let slice1_all = &mut data[..]; // Slice for the entire array
|
|
|
let ptr2_all = slice1_all.as_mut_ptr(); // Pointer for the entire array
|
|
|
|
|
|
let ptr3_at_0 = ptr2_all; // Pointer to 0th elem (the same)
|
|
|
let ptr4_at_1 = ptr2_all.add(1); // Pointer to 1th elem
|
|
|
let ref5_at_0 = &mut *ptr3_at_0; // Reference to 0th elem
|
|
|
let ref6_at_1 = &mut *ptr4_at_1; // Reference to 1th elem
|
|
|
|
|
|
*ref6_at_1 += 6;
|
|
|
*ref5_at_0 += 5;
|
|
|
*ptr4_at_1 += 4;
|
|
|
*ptr3_at_0 += 3;
|
|
|
|
|
|
// 在循环中修改所有元素( 仅仅为了有趣 )
|
|
|
// (可以使用任何裸指针，它们共享同一个借用!)
|
|
|
for idx in 0..10 {
|
|
|
*ptr2_all.add(idx) += idx;
|
|
|
}
|
|
|
|
|
|
// 同样为了有趣，再实现下安全版本的循环
|
|
|
for (idx, elem_ref) in slice1_all.iter_mut().enumerate() {
|
|
|
*elem_ref += idx;
|
|
|
}
|
|
|
|
|
|
// Should be [8, 12, 4, 6, 8, 10, 12, 14, 16, 18]
|
|
|
println!("{:?}", &data[..]);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
[8, 12, 4, 6, 8, 10, 12, 14, 16, 18]
|
|
|
|
|
|
MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
[8, 12, 4, 6, 8, 10, 12, 14, 16, 18]
|
|
|
</code></pre>
|
|
|
<h2 id="测试不可变引用"><a class="header" href="#测试不可变引用">测试不可变引用</a></h2>
|
|
|
<p>在之前的例子中，我们使用的都是可变引用，而 Rust 中还有不可变引用。那么它将如何工作呢？</p>
|
|
|
<p>我们已经见过裸指针可以被简单的拷贝只要它们共享同一个借用，那不可变引用是不是也可以这么做？</p>
|
|
|
<p>注意，下面的 <code>println</code> 会自动对待打印的目标值进行 <code>ref/deref</code> 等操作，因此为了保证测试的正确性，我们将其放入一个函数中。</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>fn opaque_read(val: &i32) {
|
|
|
println!("{}", val);
|
|
|
}
|
|
|
|
|
|
unsafe {
|
|
|
let mut data = 10;
|
|
|
let mref1 = &mut data;
|
|
|
let sref2 = &mref1;
|
|
|
let sref3 = sref2;
|
|
|
let sref4 = &*sref2;
|
|
|
|
|
|
// Random hash of shared reference reads
|
|
|
opaque_read(sref3);
|
|
|
opaque_read(sref2);
|
|
|
opaque_read(sref4);
|
|
|
opaque_read(sref2);
|
|
|
opaque_read(sref3);
|
|
|
|
|
|
*mref1 += 1;
|
|
|
|
|
|
opaque_read(&data);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
warning: unnecessary `unsafe` block
|
|
|
--> src\main.rs:6:1
|
|
|
|
|
|
|
6 | unsafe {
|
|
|
| ^^^^^^ unnecessary `unsafe` block
|
|
|
|
|
|
|
= note: `#[warn(unused_unsafe)]` on by default
|
|
|
|
|
|
warning: `miri-sandbox` (bin "miri-sandbox") generated 1 warning
|
|
|
|
|
|
10
|
|
|
10
|
|
|
10
|
|
|
10
|
|
|
10
|
|
|
11
|
|
|
</code></pre>
|
|
|
<p>虽然这里没有使用裸指针，但是可以看到对于不可变引用而言，上面的使用方式不存在任何问题。下面来增加一些裸指针：</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>fn opaque_read(val: &i32) {
|
|
|
println!("{}", val);
|
|
|
}
|
|
|
|
|
|
unsafe {
|
|
|
let mut data = 10;
|
|
|
let mref1 = &mut data;
|
|
|
let ptr2 = mref1 as *mut i32;
|
|
|
let sref3 = &*mref1;
|
|
|
let ptr4 = sref3 as *mut i32;
|
|
|
|
|
|
*ptr4 += 4;
|
|
|
opaque_read(sref3);
|
|
|
*ptr2 += 2;
|
|
|
*mref1 += 1;
|
|
|
|
|
|
opaque_read(&data);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
error[E0606]: casting `&i32` as `*mut i32` is invalid
|
|
|
--> src/main.rs:11:20
|
|
|
|
|
|
|
11 | let ptr4 = sref3 as *mut i32;
|
|
|
| ^^^^^^^^^^^^^^^^^ ^^^^^^^^^^^^^^^^^
|
|
|
</code></pre>
|
|
|
<p>可以看出，我们无法将一个不可变的引用转换成可变的裸指针，只能曲线救国了：</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>let ptr4 = sref3 as *const i32 as *mut i32;
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<p>如上，先将不可变引用转换成不可变的裸指针，然后再转换成可变的裸指针。</p>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
14
|
|
|
17
|
|
|
</code></pre>
|
|
|
<p>编译器又一次满意了，再来看看 miri :</p>
|
|
|
<pre><code class="language-shell">MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
|
|
|
error: Undefined Behavior: no item granting write access to
|
|
|
tag <1621> at alloc742 found in borrow stack.
|
|
|
--> src\main.rs:13:5
|
|
|
|
|
|
|
13 | *ptr4 += 4;
|
|
|
| ^^^^^^^^^^ no item granting write access to tag <1621>
|
|
|
| at alloc742 found in borrow stack.
|
|
|
</code></pre>
|
|
|
<p>果然，miri 提示了，原因是当我们使用不可变引用时，就相当于承诺不会去修改其中的值，那 miri 发现了这种修改行为，自然会给予相应的提示。</p>
|
|
|
<p>对此，可以用一句话来简单总结：<strong>在借用栈中，一个不可变引用，它上面的所有引用( 在它之后被推入借用栈的引用 )都只能拥有只读的权限。</strong></p>
|
|
|
<p>但是我们可以这样做:</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>fn opaque_read(val: &i32) {
|
|
|
println!("{}", val);
|
|
|
}
|
|
|
|
|
|
unsafe {
|
|
|
let mut data = 10;
|
|
|
let mref1 = &mut data;
|
|
|
let ptr2 = mref1 as *mut i32;
|
|
|
let sref3 = &*mref1;
|
|
|
let ptr4 = sref3 as *const i32 as *mut i32;
|
|
|
|
|
|
opaque_read(&*ptr4);
|
|
|
opaque_read(sref3);
|
|
|
*ptr2 += 2;
|
|
|
*mref1 += 1;
|
|
|
|
|
|
opaque_read(&data);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<p>可以看到，我们其实可以创建一个可变的裸指针，只要不去使用写操作，而是只使用读操作。</p>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
10
|
|
|
10
|
|
|
13
|
|
|
|
|
|
MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
10
|
|
|
10
|
|
|
13
|
|
|
</code></pre>
|
|
|
<p>再来检查下不可变的引用是否可以像平时一样正常弹出:</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>fn opaque_read(val: &i32) {
|
|
|
println!("{}", val);
|
|
|
}
|
|
|
|
|
|
unsafe {
|
|
|
let mut data = 10;
|
|
|
let mref1 = &mut data;
|
|
|
let ptr2 = mref1 as *mut i32;
|
|
|
let sref3 = &*mref1;
|
|
|
|
|
|
*ptr2 += 2;
|
|
|
opaque_read(sref3); // Read in the wrong order?
|
|
|
*mref1 += 1;
|
|
|
|
|
|
opaque_read(&data);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
12
|
|
|
13
|
|
|
|
|
|
MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
|
|
|
error: Undefined Behavior: trying to reborrow for SharedReadOnly
|
|
|
at alloc742, but parent tag <1620> does not have an appropriate
|
|
|
item in the borrow stack
|
|
|
|
|
|
--> src\main.rs:13:17
|
|
|
|
|
|
|
13 | opaque_read(sref3); // Read in the wrong order?
|
|
|
| ^^^^^ trying to reborrow for SharedReadOnly
|
|
|
| at alloc742, but parent tag <1620>
|
|
|
| does not have an appropriate item
|
|
|
| in the borrow stack
|
|
|
|
|
|
|
</code></pre>
|
|
|
<p>细心的同学可能会发现，我们这次获得了一个相当具体的 miri 提示，而不是之前的某个 tag 。真是令人感动...毕竟这种错误信息会更有帮助。</p>
|
|
|
<h2 id="测试内部可变性"><a class="header" href="#测试内部可变性">测试内部可变性</a></h2>
|
|
|
<p>还记得之前我们试图用 <code>RefCell</code> + <code>Rc</code> 去实现的那个糟糕的链表吗？这两个组合在一起就可以实现内部可变性。与 <code>RefCell</code> 类似的还有 <a href="https://course.rs/advance/smart-pointer/cell-refcell.html#cell"><code>Cell</code></a>:</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>use std::cell::Cell;
|
|
|
|
|
|
unsafe {
|
|
|
let mut data = Cell::new(10);
|
|
|
let mref1 = &mut data;
|
|
|
let ptr2 = mref1 as *mut Cell<i32>;
|
|
|
let sref3 = &*mref1;
|
|
|
|
|
|
sref3.set(sref3.get() + 3);
|
|
|
(*ptr2).set((*ptr2).get() + 2);
|
|
|
mref1.set(mref1.get() + 1);
|
|
|
|
|
|
println!("{}", data.get());
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<p>地狱一般的代码，就等着 miri 来优化你吧。</p>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
16
|
|
|
|
|
|
MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
16
|
|
|
</code></pre>
|
|
|
<p>等等，竟然没有任何问题，我们需要深入调查下原因:</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>pub struct Cell<T: ?Sized> {
|
|
|
value: UnsafeCell<T>,
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<p>以上是标准库中的 <code>Cell</code> 源码，可以看到里面有一个 <code>UnsafeCell</code>，通过名字都能猜到，这个数据结构相当的不安全，在<a href="https://doc.rust-lang.org/std/cell/struct.UnsafeCell.html">标准库</a>中有以下描述:</p>
|
|
|
<blockquote>
|
|
|
<p>Rust 中用于内部可变性的核心原语( primitive )。</p>
|
|
|
<p>如果你拥有一个引用 <code>&T</code>，那一般情况下, Rust编译器会基于 <code>&T</code> 指向不可变的数据这一事实来进行相关的优化。通过别名或者将 <code>&T</code> 强制转换成 <code>&mut T</code> 是一种 UB 行为。</p>
|
|
|
<p>而 <code>UnsafeCell<T></code> 移除了 <code>&T</code> 的不可变保证：一个不可变引用 <code>&UnsafeCell<T></code> 指向一个可以改变的数据。，这就是内部可变性。</p>
|
|
|
</blockquote>
|
|
|
<p>感觉像是魔法，那下面就用该魔法让 miri happy 下:</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>use std::cell::UnsafeCell;
|
|
|
|
|
|
fn opaque_read(val: &i32) {
|
|
|
println!("{}", val);
|
|
|
}
|
|
|
|
|
|
unsafe {
|
|
|
let mut data = UnsafeCell::new(10);
|
|
|
let mref1 = &mut data; // Mutable ref to the *outside*
|
|
|
let ptr2 = mref1.get(); // Get a raw pointer to the insides
|
|
|
let sref3 = &*mref1; // Get a shared ref to the *outside*
|
|
|
|
|
|
*ptr2 += 2; // Mutate with the raw pointer
|
|
|
opaque_read(&*sref3.get()); // Read from the shared ref
|
|
|
*sref3.get() += 3; // Write through the shared ref
|
|
|
*mref1.get() += 1; // Mutate with the mutable ref
|
|
|
|
|
|
println!("{}", *data.get());
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
12
|
|
|
16
|
|
|
|
|
|
MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
12
|
|
|
16
|
|
|
</code></pre>
|
|
|
<p>这段代码非常成功！但是等等..这里的代码顺序有问题：我们首先获取了内部的裸指针 <code>ptr2</code>，然后获取了一个不可变引用 <code>sref3</code>，接着我们使用了裸指针，然后是 <code>sref3</code>，这不就是标准的借用栈错误典范吗？既然如此，为何 miri 没有给出提示？</p>
|
|
|
<p>现在有两个解释:</p>
|
|
|
<ul>
|
|
|
<li>Miri 并不完美，它依然会有所遗漏，也会误判</li>
|
|
|
<li>我们的简化模型貌似过于简化了</li>
|
|
|
</ul>
|
|
|
<p>大家选择哪个？..我不管，反正我选择第二个。不过，虽然我们的借用栈过于简单，但是依然是亲孩子嘛，最后再基于它来实现一个真正正确的版本:</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>use std::cell::UnsafeCell;
|
|
|
|
|
|
fn opaque_read(val: &i32) {
|
|
|
println!("{}", val);
|
|
|
}
|
|
|
|
|
|
unsafe {
|
|
|
let mut data = UnsafeCell::new(10);
|
|
|
let mref1 = &mut data;
|
|
|
// These two are swapped so the borrows are *definitely* totally stacked
|
|
|
let sref2 = &*mref1;
|
|
|
// Derive the ptr from the shared ref to be super safe!
|
|
|
let ptr3 = sref2.get();
|
|
|
|
|
|
*ptr3 += 3;
|
|
|
opaque_read(&*sref2.get());
|
|
|
*sref2.get() += 2;
|
|
|
*mref1.get() += 1;
|
|
|
|
|
|
println!("{}", *data.get());
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
13
|
|
|
16
|
|
|
|
|
|
MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
13
|
|
|
16
|
|
|
</code></pre>
|
|
|
<h2 id="测试-box"><a class="header" href="#测试-box">测试 Box</a></h2>
|
|
|
<p>大家还记得为何我们讲了这么长的两章借用栈吗？原因就在于 <code>Box</code> 和裸指针混合使用时出了问题。</p>
|
|
|
<p><code>Box</code> 在某种程度上类似 <code>&mut</code>，因为对于它指向的内存区域，它拥有唯一的所有权。</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>unsafe {
|
|
|
let mut data = Box::new(10);
|
|
|
let ptr1 = (&mut *data) as *mut i32;
|
|
|
|
|
|
*data += 10;
|
|
|
*ptr1 += 1;
|
|
|
|
|
|
// Should be 21
|
|
|
println!("{}", data);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
21
|
|
|
|
|
|
MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
|
|
|
error: Undefined Behavior: no item granting read access
|
|
|
to tag <1707> at alloc763 found in borrow stack.
|
|
|
|
|
|
--> src\main.rs:7:5
|
|
|
|
|
|
|
7 | *ptr1 += 1;
|
|
|
| ^^^^^^^^^^ no item granting read access to tag <1707>
|
|
|
| at alloc763 found in borrow stack.
|
|
|
|
|
|
|
</code></pre>
|
|
|
<p>现在到现在为止，大家一眼就能看出来这种代码不符合借用栈的规则。当然, miri 也讨厌这一点，因此我们来改正下。</p>
|
|
|
<pre><pre class="playground"><code class="language-rust edition2021"><span class="boring">#![allow(unused)]
|
|
|
</span><span class="boring">fn main() {
|
|
|
</span>unsafe {
|
|
|
let mut data = Box::new(10);
|
|
|
let ptr1 = (&mut *data) as *mut i32;
|
|
|
|
|
|
*ptr1 += 1;
|
|
|
*data += 10;
|
|
|
|
|
|
// Should be 21
|
|
|
println!("{}", data);
|
|
|
}
|
|
|
<span class="boring">}</span></code></pre></pre>
|
|
|
<pre><code class="language-shell">$ cargo run
|
|
|
|
|
|
21
|
|
|
|
|
|
MIRIFLAGS="-Zmiri-tag-raw-pointers" cargo +nightly-2022-01-21 miri run
|
|
|
21
|
|
|
</code></pre>
|
|
|
<p>在经过这么长的旅程后，我们终于完成了借用栈的学习，兄弟们我已经累趴了，你们呢？</p>
|
|
|
<p>但是，话说回来，该如何使用 <code>Box</code> 来解决栈借用的问题？当然，我们可以像之前的测试例子一样写一些玩具代码，但是在实际链表中中，将 <code>Box</code> 存储在某个地方，然后长时间持有一个裸指针才是经常遇到的。</p>
|
|
|
<p>等等，你说链表？天呐，我都忘记了我们还在学习链表，那接下来，继续实现之前未完成的链表吧。</p>
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