我正在遵循此示例，以将Rust写入python App的函数导入： https://depth-first.com/articles/2022/03/09/python-extensions-in-python-extension in
输出是'libfunction.dylib'，我应该将其重命名为'function.so'，将其移至文件夹中，然后我应该能够在python中导入它（例如该模块只是公开一个称为entry的函数）：
from function import greet
我缺少的python环境设置了一些设置吗？即使我完全使用文章链接的存储库，python也找不到模块（检查了正确拼写的10次，该文件在正确的文件夹中等等。）：
ModuleNotFoundError: No module named function
货物
[package]
name = "function"
version = "0.1.0"
edition = "2021"
[lib]
crate-type = ["cdylib"]
[dependencies]
cpython = "0.7"
[features]
default = ["python3"]
python3 = ["cpython/python3-sys", "cpython/extension-module"]
。 .rs：
use cpython::{py_module_initializer, py_fn, PyResult, Python};
fn greet(_: Python, name: String) -> PyResult<String> {
Ok(format!("Hello, {}!", name))
}
py_module_initializer! {
function, |py, module| {
module.add(py, "greet", py_fn!(py, greet(string: String)))?;
Ok(())
}
}
I'm following through this example for importing a function written in Rust into a python app: https://depth-first.com/articles/2022/03/09/python-extensions-in-pure-rust-with-rust-cpython/
the output is 'libfunction.dylib', which I should rename to 'function.so', move it to the folder where there's the python app.py, then I should be able to import it in python like (say that the module is just exposing a function called greet):
from function import greet
Is there some setup of the python environment that I'm missing? Even when I use exactly the repository linked by the article, python can't find the module (checked 10 times that the names are spelled correctly, the file is in the right folder etc..):
ModuleNotFoundError: No module named function
The cargo.toml would be
[package]
name = "function"
version = "0.1.0"
edition = "2021"
[lib]
crate-type = ["cdylib"]
[dependencies]
cpython = "0.7"
[features]
default = ["python3"]
python3 = ["cpython/python3-sys", "cpython/extension-module"]
the lib.rs:
use cpython::{py_module_initializer, py_fn, PyResult, Python};
fn greet(_: Python, name: String) -> PyResult<String> {
Ok(format!("Hello, {}!", name))
}
py_module_initializer! {
function, |py, module| {
module.add(py, "greet", py_fn!(py, greet(string: String)))?;
Ok(())
}
}
发布评论
评论（1）
如 py_module_initializer！{function，...} macro生成一个名为
pyinit_function的模块导出函数，共享对象名称必须与之匹配。重命名
libFunction.so函数和类似地导入：As suggested in this issue,
the
py_module_initializer!{function, ...}macro generates a module export function namedPyInit_function, and the shared object name must match it.Rename
libfunction.sotofunction.soand import like this: