在 Rust 中将原始指针转换为 16 位 Unicode 字符到文件路径 [英] Converting raw pointer to 16-bit Unicode character to file path in Rust

问题描述

我正在用一个用 Rust 编写的 DLL 替换一个用 C++ 编写的 DLL.目前DLL中的函数调用如下:

I'm replacing a DLL written in C++ with one written in Rust. Currently the function in the DLL is called as follows:

BOOL calledFunction(wchar_t* pFileName)

我相信在这种情况下 wchar_t 是一个 16 位 Unicode 字符，所以我选择在我的 Rust DLL 中公开以下函数:

I believe that in this context wchar_t is a 16-bit Unicode character, so I chose to expose the following function in my Rust DLL:

pub fn calledFunction(pFileName: *const u16)

将原始指针转换为我实际上可以用来从 Rust DLL 打开文件的内容的最佳方法是什么?

What would be the best way to convert that raw pointer to something I could actually use to open the file from the Rust DLL?

推荐答案

以下是一些示例代码:

use std::ffi::OsString;
use std::os::windows::prelude::*;

unsafe fn u16_ptr_to_string(ptr: *const u16) -> OsString {
    let len = (0..).take_while(|&i| *ptr.offset(i) != 0).count();
    let slice = std::slice::from_raw_parts(ptr, len);

    OsString::from_wide(slice)
}

// main example
fn main() {
    let buf = vec![97_u16, 98, 99, 100, 101, 102, 0];
    let ptr = buf.as_ptr(); // raw pointer

    let string = unsafe { u16_ptr_to_string(ptr) };

    println!("{:?}", string);
}

在u16_ptr_to_string中，你做了三件事:

• 通过使用 偏移(不安全)
• 使用 from_raw_parts<创建切片/a>(不安全)
• 使用 from_wide

• get the length of the string by counting the non-zero characters using offset (unsafe)
• create a slice using from_raw_parts (unsafe)
• transform this &[u16] into an OsString with from_wide

最好使用 wchar_t 和 wcslen 来自 libc 板条箱并使用另一个板条箱进行转换.重新实现已经在 crate 中维护的东西可能是个坏主意.

It is better to use wchar_t and wcslen from the libc crate and use another crate for conversion. This is maybe a bad idea to reimplement something that is already maintained in a crate.

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