Question

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"]

我构建它以获得target/release/libembed.so，并在此路径中创建invoke.py：

from ctypes import cdll

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

lib.process()

print("done!")

执行，错误：

-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

如何解决这个问题？

Answer 1

#[no_mangle]继续使用该功能，而不是use。

#[no_mangle]
pub extern fn process() { ... }

有关完整的工作示例，请参阅the Rust FFI Omnibus "Integers" example。

从Python调用Rust时出错

1 个答案: