type OnProduce = extern "C" fn 不是 FFI 安全的

Question

pub type OnProduce = extern "C" fn(*mut ZLMedia, *const u8, size_t);
extern "C" {
    pub fn zlmedia_set_on_produce(zl_media: *mut ZLInstance, on_produce: OnProduce);
}

I get:

   |
23 |     pub fn zlmedia_set_on_produce(zl_media: *mut ZLInstance, on_produce: OnProduce);
   |                                                                          ^^^^^^^^^ not FFI-safe
   |
   = help: consider adding a `#[repr(C)]` or `#[repr(transparent)]` attribute to this struct
   = note: this struct has unspecified layout

but it's not possible to add #[repr(C)] for a type, only structs. As you can see, OnProduce is an extern "C" function. I thought it'd be already FFI-safe

Answer 1

As @Frxstream pointed out, it's probably because ZLMedia isn't FFI-safe.

Quote:

The trouble is that in Rust when a struct is repr(Rust) (the default) it is then free to use whatever layout it thinks is most efficient for your program. So this could (hypothetically) be different between two Rust programs and is even more likely to be different if the two programs are compiled with different versions of the Rust compiler.

So for FFI you need an explicitly specified layout. Currently C and transparent are the only stable layouts for structs. There may be more more in the future but that's currently a ways off.

I recently had a discussion about this on the forums, which you might be interested to read (origin of the quote): [link]