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Merge pull request #37 from ch3pjw/pin_fixes

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Pin fixes
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Carl Fredrik Samson
2021-04-19 14:01:01 +02:00
committed by GitHub
parent 9da948ef1d 156ae48bbf
commit 3def942a62
1 changed files with 5 additions and 6 deletions
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src/5_pin.md
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@@ -35,7 +35,7 @@ for the names that were chosen. Naming is not easy, and I considered renaming
`Unpin` and `!Unpin` in this book to make them easier to reason about. `Unpin` and `!Unpin` in this book to make them easier to reason about.
However, an experienced member of the Rust community convinced me that that there However, an experienced member of the Rust community convinced me that that there
is just too many nuances and edge-cases to consider which are easily overlooked when are just too many nuances and edge-cases to consider which are easily overlooked when
naively giving these markers different names, and I'm convinced that we'll naively giving these markers different names, and I'm convinced that we'll
just have to get used to them and use them as is. just have to get used to them and use them as is.
@@ -213,7 +213,7 @@ now. The struct is not self-referential anymore, it holds a pointer to a field
in a different object. That means we can't rely on the lifetime of `test2.b` to in a different object. That means we can't rely on the lifetime of `test2.b` to
be tied to the lifetime of `test2` anymore. be tied to the lifetime of `test2` anymore.
If your still not convinced, this should at least convince you: If you're still not convinced, this should at least convince you:
```rust ```rust
fn main() { fn main() {
@@ -289,7 +289,6 @@ struct Test {
impl Test { impl Test {
fn new(txt: &str) -> Self { fn new(txt: &str) -> Self {
let a = String::from(txt);
Test { Test {
a: String::from(txt), a: String::from(txt),
b: std::ptr::null(), b: std::ptr::null(),
@@ -528,8 +527,8 @@ pub fn main() {
let mut test1 = Test::new("test1"); let mut test1 = Test::new("test1");
let mut test2 = Test::new("test2"); let mut test2 = Test::new("test2");
println!("a: {}, b: {}",test1.as_ref().a(), test1.as_ref().b()); println!("a: {}, b: {}", test1.as_ref().a(), test1.as_ref().b());
println!("a: {}, b: {}",test2.as_ref().a(), test2.as_ref().b()); println!("a: {}, b: {}", test2.as_ref().a(), test2.as_ref().b());
} }
``` ```
@@ -550,7 +549,7 @@ to be moved even when pinned, so `Pin` will have no effect on such a type.
2. Getting a `&mut T` to a pinned T requires unsafe if `T: !Unpin`. In 2. Getting a `&mut T` to a pinned T requires unsafe if `T: !Unpin`. In
other words: requiring a pinned pointer to a type which is `!Unpin` prevents other words: requiring a pinned pointer to a type which is `!Unpin` prevents
the _user_ of that API from moving that value unless it choses to write `unsafe` the _user_ of that API from moving that value unless they choose to write `unsafe`
code. code.
3. Pinning does nothing special with memory allocation like putting it into some 3. Pinning does nothing special with memory allocation like putting it into some