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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 >
< 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 >
< 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， ， < / 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 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 >
< 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 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 > < 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 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 > < 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 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 > < 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 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 > < 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 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 > < 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 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 > < 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， < / p >
< 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 > < 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， < code > .split_at_mut(position)< / code > 来将一个数组分成多个部分:< / p >
< 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 > < 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 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 > < 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 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 > < 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 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 > < 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 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 >
< 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， < / 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 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 >
< 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 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 > < 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 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 >
< 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 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 >
< 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 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 > < 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 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 > < 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 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 > < 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 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 > < 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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