Fix opaque type representation.

opaque-type-auto-traits
Florian Gilcher 4 years ago
parent a5a48441d4
commit 882b491288

@ -740,8 +740,18 @@ void bar(struct Bar *arg);
To do this in Rust, lets create our own opaque types: To do this in Rust, lets create our own opaque types:
```rust ```rust
#[repr(C)] pub struct Foo { _private: [u8; 0] } #[repr(C)]
#[repr(C)] pub struct Bar { _private: [u8; 0] } pub struct Foo {
_data: [u8; 0],
_marker:
core::marker::PhantomData<(*mut u8, core::marker::PhantomPinned)>,
}
#[repr(C)]
pub struct Bar {
_data: [u8; 0],
_marker:
core::marker::PhantomData<(*mut u8, core::marker::PhantomPinned)>,
}
extern "C" { extern "C" {
pub fn foo(arg: *mut Foo); pub fn foo(arg: *mut Foo);
@ -750,12 +760,12 @@ extern "C" {
# fn main() {} # fn main() {}
``` ```
By including a private field and no constructor, By including at least one private field and no constructor,
we create an opaque type that we can't instantiate outside of this module. we create an opaque type that we can't instantiate outside of this module.
(A struct with no field could be instantiated by anyone.) (A struct with no field could be instantiated by anyone.)
We also want to use this type in FFI, so we have to add `#[repr(C)]`. We also want to use this type in FFI, so we have to add `#[repr(C)]`.
And to avoid warning around using `()` in FFI, we instead use an empty array, The marker ensures the compiler does not mark the struct as `Send`, `Sync` and `Unpin` are
which works just as well as an empty type but is FFI-compatible. not applied to the struct. (`*mut u8` is not `Send` or `Sync`, `PhantomPinned` is not `Unpin`)
But because our `Foo` and `Bar` types are But because our `Foo` and `Bar` types are
different, well get type safety between the two of them, so we cannot different, well get type safety between the two of them, so we cannot

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