How to expose a Rust `Vec<T>` to FFI?

25
votes

I'm trying to construct a pair of elements:

  • array: *mut T
  • array_len: usize

array is intended to own the data

However, Box::into_raw will return *mut [T]. I cannot find any info on converting raw pointers to slices. What is its layout in memory? How do I use it from C? Should I convert to *mut T? If so, how?

2
rustffi

2 Answers

25
votes

If you just want some C function to mutably borrow the Vec, you can do it like this:

extern "C" {
    fn some_c_function(ptr: *mut i32, len: ffi::size_t);
}

fn safe_wrapper(a: &mut [i32]) {
    unsafe {
        some_c_function(a.as_mut_ptr(), a.len() as ffi::size_t);
    }
}

Of course, the C function shouldn't store this pointer somewhere else because that would break aliasing assumptions.

If you want to "pass ownership" of the data to C code, you'd do something like this:

use std::mem;

extern "C" {
    fn c_sink(ptr: *mut i32, len: ffi::size_t);
}

fn sink_wrapper(mut vec: Vec<i32>) {
    vec.shrink_to_fit();
    assert!(vec.len() == vec.capacity());
    let ptr = vec.as_mut_ptr();
    let len = vec.len();
    mem::forget(vec); // prevent deallocation in Rust
                      // The array is still there but no Rust object
                      // feels responsible. We only have ptr/len now
                      // to reach it.
    unsafe {
        c_sink(ptr, len as ffi::size_t);
    }
}

Here, the C function "takes ownership" in the sense that we expect it to eventually return the pointer and length to Rust, for example, by calling a Rust function to deallocate it:

#[no_mangle]
/// This is intended for the C code to call for deallocating the
/// Rust-allocated i32 array.
unsafe extern "C" fn deallocate_rust_buffer(ptr: *mut i32, len: ffi::size_t) {
    let len = len as usize;
    drop(Vec::from_raw_parts(ptr, len, len));
}

Because Vec::from_raw_parts expects three parameters, a pointer, a size and a capacity, we either have to keep track of the capacity as well somehow, or we use Vec's shrink_to_fit before passing the pointer and length to the C function. This might involve a reallocation, though.

10
votes

You could use [T]::as_mut_ptr to obtain the *mut T pointer directly from Vec<T>, Box<[T]> or any other DerefMut-to-slice types.

use std::mem;

let mut boxed_slice: Box<[T]> = vector.into_boxed_slice();

let array: *mut T = boxed_slice.as_mut_ptr();
let array_len: usize = boxed_slice.len();

// Prevent the slice from being destroyed (Leak the memory).
mem::forget(boxed_slice);