如何在 Rust 和 Python 之间传递指针？

Question

我正在尝试用 Rust 编写一个库，我可以从 Python 代码中调用它。我希望能够将一个 void 指针传回 Python，这样我就可以在 Rust 调用之间保持状态。但是，当我再次尝试访问指针时，我在 Rust 中遇到了段错误。

完整代码示例和崩溃报告：https://gist.github.com/robyoung/3644f13a05c95cb1b947

密码

#![feature(libc)]
#![feature(alloc)]
extern crate libc;
use std::boxed;

pub struct Point {
    x: i64,
    y: i32,
}

#[no_mangle]
pub extern "C" fn start_state() -> *mut Point {
    let point = Box::new(Point{x: 0, y: 10});
    let raw = unsafe { boxed::into_raw(point) };
    println!("{:?}", raw);
    raw
}

#[no_mangle]
pub extern "C" fn continue_state(point: *mut Point) -> i32 {
    println!("{:?}", point);
    let p = unsafe { Box::from_raw(point) };
    println!("{} {}", p.x, p.y);
    0
}

import ctypes

lib = ctypes.cdll.LoadLibrary('target/libpytesttype.so')
lib.start_state.restype = ctypes.c_void_p

pointer = lib.start_state()
print("{:x}".format(pointer))
lib.continue_state(pointer)

输出

0xdc24000
10dc24000
0xdc24000
[1]    64006 segmentation fault  python src/main.py

我做错了什么？

Answer 1

:

On the Python side, you're missing lib.continue_state.argtypes = (ctypes.c_void_p,). Without defining the parameter as a pointer, ctypes uses the default conversion for a Python integer, which truncates the value to 32-bit, e.g. 0x0dc24000. If you're lucky accessing that address triggers a segfault immediately.

我的输出（带有我自己的填充）是：

0x103424000
  103424000
0x  3424000

所以指针的 Debug 格式化程序应该没问题。不确定为什么你的输出不同。

添加后

lib.continue_state.argtypes = (ctypes.c_void_p,)

程序运行就好了。

如何在 Rust 和 Python 之间传递指针？

How can a pointer be passed between Rust and Python?

python

ctypes

rust

密码

输出