Python调用Rust编译的共享库出现错误

libraco
  • 760

Rust Code:

#[no_mangle]
use std::thread;

pub extern fn process() {
    let handles: Vec<_> = (0..10).map(|_| {
        thread::spawn(|| {
            let mut _x = 0;
            for _ in (0..5_000_001) {
                _x += 1
            }
        })
    }).collect();

    for h in handles {
        h.join().ok().expect("Could not join a thread!");
    }
}

Cargo.toml:

[package]
name = "embed"
version = "0.1.0"
authors = ["hustlibraco <hustlibraco@gmail.com>"]

[lib]
name = "embed"
crate-type = ["dylib"]

编译得到target/release/libembed.so共享库,然后我在此目录下面新建了一个python文件:

from ctypes import cdll

lib = cdll.LoadLibrary("target/release/libembed.so")

lib.process()

print("done!")

执行报错:

-bash-4.2# python invoke.py 
Traceback (most recent call last):
  File "invoke.py", line 5, in <module>
    lib.process()
  File "/usr/lib64/python2.7/ctypes/__init__.py", line 373, in __getattr__
    func = self.__getitem__(name)
  File "/usr/lib64/python2.7/ctypes/__init__.py", line 378, in __getitem__
    func = self._FuncPtr((name_or_ordinal, self))
AttributeError: target/release/libembed.so: undefined symbol: process

这是Rust官方教程给的例子,为什么我这里会执行出错呢?

回复
阅读 4.5k
2 个回答
✓ 已被采纳

#[no_mangle]应该写在函数前面,而不是use前...

#[no_mangle]
pub extern fn process() {···}
Under_The_Rose
  • 1
新手上路,请多包涵

楼主你是怎么编译的,我编译出来的结果没有libembed.so,好捉急_(:з」∠)_

撰写回答
你尚未登录,登录后可以
  • 和开发者交流问题的细节
  • 关注并接收问题和回答的更新提醒
  • 参与内容的编辑和改进,让解决方法与时俱进
宣传栏