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How to lend a Rust object to C code for an arbitrary lifetime?

Tags:

c

rust

ffi

I'm writing a library in Rust that has a C interface. C side must be able to create and destroy Rust objects (C side owns them and controls their lifetime).

I've managed to "leak" an object to C, but I'm not sure how to properly free it:

pub extern "C" fn create() -> *mut Foo {
   let obj = Foo; // oops, a bug
   let ptr = std::mem::transmute(&mut obj); // bad 
   std::mem::forget(obj); // not needed
   return ptr;
}

pub extern "C" fn destroy(handle: *mut Foo) {
   // get Foo back and Drop it??? 
}

I'm not sure how can I turn pointer back to an object that Rust will call Drop on. Simply dereferencing *handle doesn't compile.

like image 889
Kornel Avatar asked Feb 02 '15 12:02

Kornel


2 Answers

To send a Rust object to C:

#[no_mangle]
pub extern "C" fn create_foo() -> *mut Foo {
    Box::into_raw(Box::new(Foo))
}

or taking advantage of Box being FFI-safe and the same as a pointer, and the fact that Rust function definitions do not have to match C headers exactly, as long as the ABI is the same:

#[no_mangle]
pub extern "C" fn create_foo() -> Box<Foo> {
   Box::new(Foo)
}

(returning Option<Box<Foo>> is fine too. Result is not.)

To borrow (and not free) from C:

#[no_mangle]
pub unsafe extern "C" fn peek_at(foo: *mut Foo) {
    let foo = foo.as_ref().unwrap(); // That's ptr::as_ref
}

or taking advantage of references and Option being FFI-safe:

#[no_mangle]
pub extern "C" fn peek_at(foo: Option<&mut Foo>) {
    let foo = foo.unwrap();
}

To take over/destroy Rust object previously given to C:

#[no_mangle]
pub unsafe extern "C" fn free_foo(foo: *mut Foo) {
    assert!(!foo.is_null());
    Box::from_raw(foo); // Rust auto-drops it
}

or using the fact that Option<Box> is FFI-safe, and memory-managed by Rust:

#[no_mangle]
pub unsafe extern "C" fn free_foo(foo: Option<Box<Foo>>) {
   // dropped implicitly
}
like image 88
Kornel Avatar answered Nov 07 '22 10:11

Kornel


Actually, you haven't managed to leak an object to C; you've managed to leak a reference to a (shortly) non-existent stack frame. :D

Here's a full example that should work correctly. I've tried to comment it as appropriate to explain what I'm doing and why.

pub struct Dramatic(String);

// Implement a destructor just so we can see when the object is destroyed.
impl Drop for Dramatic {
    fn drop(&mut self) {
        println!("And lo, I, {}, meet a most terrible fate!", self.0);
    }
}

pub extern "C" fn create() -> *mut Dramatic {
    // We **must** heap-allocate the object!  Returning a reference to a local
    // will **almost certainly** break your program!
    let mut obj = Box::new(Dramatic("Roger".to_string()));

    // into_raw turns the Box into a *mut Dramatic, which the borrow checker
    // ignores, without calling its destructor.
    Box::into_raw(obj)
}

pub extern "C" fn destroy(ptr: &mut *mut Dramatic) {
    // First, we **must** check to see if the pointer is null.
    if ptr.is_null() {
        // Do nothing.
        return;
    }

    // Now we know the pointer is non-null, we can continue. from_raw is the
    // inverse of into_raw: it turns the *mut Dramatic back into a
    // Box<Dramatic>. You must only call from_raw once per pointer.
    let obj: Box<Dramatic> = unsafe { Box::from_raw(*ptr) };

    // We don't *have* to do anything else; once obj goes out of scope, it will
    // be dropped.  I'm going to drop it explicitly, however, for clarity.
    drop(obj);

    // I am, however, going to null out the `ptr` we were passed just so the
    // calling code is less likely to accidentally re-use the pointer.
    *ptr = ::std::ptr::null_mut();
}

fn main() {
    let mut ptr = create();
    println!("ptr = {:?}", ptr);
    destroy(&mut ptr);
    println!("ptr = {:?}", ptr);
}
like image 10
DK. Avatar answered Nov 07 '22 11:11

DK.