Question

我有一个需要通过ctypes模块导入到Python的Rust库。我的目标是使用带有Vec<T> / i32作为参数的Rust函数，并从Python返回这些类型。目前，我可以将整数传递给Rust函数，并让它们返回列表/整数。这是当前的代码：

的Python：

import ctypes
from ctypes import cdll

class List_4(ctypes.Structure):
    _fields_ = [("array", ctypes.ARRAY(ctypes.c_int32, 4))]

rust = cdll.LoadLibrary("target/debug/py_link.dll")
rust.function_vec.restype = List_4

foobar = rust.function_i32(5)
barbaz = rust.function_vec([1, 2, 3, 4]) # Throws error: Don't know how to convert parameter

print foobar
print barbaz

锈：

#[repr(C)]
pub struct List_4 {
    array: [i32; 4]
}

#[no_mangle]
pub extern fn function_i32(number: i32) -> i32 {
    number
}

#[no_mangle]
pub extern fn function_vec(list: List_4) -> List_4 {
    List_4 { array: [1, 2, 3, 5] }
}

我需要帮助的是将Python列表作为参数传递给Rust函数。我最好的猜测是将ctypes.ARRAY传递给函数而不是列表，但我不确定如何将Python列表转换为该类型。

注意：我尝试了来自this related question的Rust代码，但是当我尝试编译它时，它说“与`gcc`链接失败：退出代码：1”和“错误的重定位地址”。

Answer 1

看起来我解决了这个问题。我将Python列表转换为C数组，并将其传递给Rust函数。这是工作代码：

#[repr(C)]
pub struct List_4 {
    // Create a struct using #[repr(C)], will do the same in Python so it is shareable
    array: [i32; 4]
}

#[no_mangle]
pub extern fn function_array(list: List_4) -> List_4 {
    // Return a new instance of List_4
    List_4 { array: [1, 2, 3, 5] }
}

的Python：

import ctypes # By using ctypes, and #[repr(C)], we use the same type
              # of data in both languages, so it is possible to send stuff between them

rust = cdll.LoadLibrary("target/debug/py_link.dll") # Import the Rust dll

class List_4(ctypes.Structure):
    # Similar to creating the struct in Rust
    _fields_ = [("array", ctypes.ARRAY(ctypes.c_int32, 4))]

rust.function_array.restype = List_4 # Set the function's return type to List_4

def function_array(arr):
    # For convenience, a function to convert a list to a C array, call the function,
    # and convert its return value to a list
    return list(
        rust.function_array(
            (ctypes.c_int * len(lst))(*lst) # Call the function after converting the list 
        ).array
    )

# To use the function:
>>> function_array([1, 2, 3])
[1, 2, 3, 5]

将Python列表传递给Rust函数

1 个答案: