与数字数组的情况完全没有区别。 C 字符串是以零结尾的字节数组,因此它们在 Rust 中的表示形式为*const c_char
,然后可以转换为&CStr
然后可以用来获得&[u8]
进而&str
.
Python:
import ctypes
rustLib = "libtest.dylib"
def testRust():
lib = ctypes.cdll.LoadLibrary(rustLib)
list_to_send = ['blah', 'blah', 'blah', 'blah']
c_array = (ctypes.c_char_p * len(list_to_send))(*list_to_send)
lib.get_strings(c_array, len(list_to_send))
if __name__=="__main__":
testRust()
Rust:
#![feature(libc)]
extern crate libc;
use std::slice;
use std::ffi::CStr;
use std::str;
use libc::{size_t, c_char};
#[no_mangle]
pub extern fn get_strings(array: *const *const c_char, length: size_t) {
let values = unsafe { slice::from_raw_parts(array, length as usize) };
let strs: Vec<&str> = values.iter()
.map(|&p| unsafe { CStr::from_ptr(p) }) // iterator of &CStr
.map(|cs| cs.to_bytes()) // iterator of &[u8]
.map(|bs| str::from_utf8(bs).unwrap()) // iterator of &str
.collect();
println!("{:?}", strs);
}
Running:
% rustc --crate-type=dylib test.rs
% python test.py
["blah", "blah", "blah", "blah"]
再说一次,你应该小心生命周期并确保Vec<&str>
不会比 Python 端的原始值更长久。