如何使用PyO3解组PyCodeObject？

Question

我正在阅读.pyc文件，需要能够解组代码对象。当我尝试将解组的PyAny向下转换为PyCodeObject时，会收到以下错误消息：

error[E0277]: the trait bound `pyo3::ffi::code::PyCodeObject: pyo3::type_object::PyTypeInfo` is not satisfied
   --> src/lib.rs:179:47
    |
179 |         let code = *(loads(py, &code_buffer)?.downcast::<PyCodeObject>()?);
    |                                               ^^^^^^^^ the trait `pyo3::type_object::PyTypeInfo` is not implemented for `pyo3::ffi::code::PyCodeObject`
    |
    = note: required because of the requirements on the impl of `for<'py> pyo3::conversion::PyTryFrom<'py>` for `pyo3::ffi::code::PyCodeObject`

error[E0277]: the trait bound `pyo3::ffi::code::PyCodeObject: pyo3::instance::PyNativeType` is not satisfied
   --> src/lib.rs:179:47
    |
179 |         let code = *(loads(py, &code_buffer)?.downcast::<PyCodeObject>()?);
    |                                               ^^^^^^^^ the trait `pyo3::instance::PyNativeType` is not implemented for `pyo3::ffi::code::PyCodeObject`
    |
    = note: required because of the requirements on the impl of `for<'py> pyo3::conversion::PyTryFrom<'py>` for `pyo3::ffi::code::PyCodeObject`

正确的方法是什么

MCVE

use pyo3::{ffi::PyCodeObject, marshal::loads, Python};

fn main() {
    let gil_guard = Python::acquire_gil();
    let py = gil_guard.python();
    let code_buffer = &include_bytes!("__pycache__/test.cpython-37.pyc")[16..];
    let code = *(loads(py, &code_buffer)
        .unwrap()
        .downcast::<PyCodeObject>()
        .unwrap());
}

要创建测试文件，请执行以下操作：

1. 创建一个.py文件
2. 用Python导入模块（例如python(3) -c 'import ...'）
3. __pycache__文件夹中应该有一个.pyc文件
4. 用实际路径替换对include_bytes!的调用中代码中的路径

---

版本信息

• Rust 2018
• rustc 1.43.0-每晚（564758c4c 2020-03-08）
• 货运1.43.0-每晚（bda50510d 2020-03-02）
• CPython 3.7.3
• PyO3 0.9.1

Answer 1

我想我已经想出了如何做到这一点：

let code_ptr = loads(py, &code_buffer)?.as_ptr() as *mut PyCodeObject;
// This should be valid, since PyCodeObject is Copy, as long as the refcount is positive
let code = unsafe { *code_ptr };