如何使用PyCodeObject解除PyO3的编组？

Question

我正在读取.pyc文件，需要能够解组代码对象。当我试图将已解封的PyAny降为PyCodeObject时，我会得到以下错误消息：

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`

做这件事的正确方法是什么？

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：

python(3) -c 'import ...')

• There folder

• Replace中的

1. 创建.py文件
2. 导入模块(例如，应该是__pycache__
   1. 中的.pyc文件，在调用include_bytes!的代码中使用实际路径

   中的路径)

版本信息

2020-03-08)

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

  )

Answer 1

我想我已经想好了该怎么做

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 };