rust - 如何从C++填充Rust函数指针

Question

我想从Rust调用C++。然后，C++分配并填充缓冲区，然后Rust使用它并取消分配。但是，我想返回函数指针以取消分配数据。这是我尝试的:
在Rust中:

extern "C" {
    pub fn openvpn_receive(
        instance: *mut OpenVpnInstance,
        size: *mut size_t,
        deallocate: extern "C" fn(*mut u8),
    ) -> *mut u8;
}

fn main() {
    let size: *mut size_t;
    let deallocate = extern "C" fn(*mut u8);
    let data: *mut u8 = openvpn_receive(self.instance, size, deallocate);
    //use data
    deallocate(data);
}

在C++中:

uint8_t* openvpn_receive(size_t *size, void (*deallocate_function)(uint8_t *))
{
    deallocate_function = &cpp_deallocate_u8;
    uint8*t data = receive_data(size);
}

但是Rust代码无效。我该如何使这种事情发生？

Answer 1

也许会对您有帮助。
https://doc.rust-lang.org/std/keyword.extern.html
https://rust-embedded.github.io/book/interoperability/rust-with-c.html
C++-> C-> rust

use libc::{size_t, c_void};

#[repr(C)]
pub struct OpenVpnInstance();

//#[link(name = "your_c_library")]
extern "C" {
    pub fn openvpn_receive(
        instance: *mut OpenVpnInstance,
        size: *mut size_t,
        deallocate: *mut c_void,
    ) -> *mut u8;
}

pub fn my_callback(arg: *mut u8)
{
    println!("CB!");
}

fn main() 
{
    let size: size_t = 42;
    let instance = OpenVpnInstance{};
    
    let dummy_arg: u8 = 42;
    
    unsafe {
        let cb = my_callback as *mut c_void;
        let instance_ptr = &instance as *const _ as *mut OpenVpnInstance;
        let arg_ptr = &dummy_arg as *const _ as *mut u8;
        openvpn_receive(instance_ptr, size as *mut size_t, cb);
        my_callback(arg_ptr)
    }
}