如何将某些堆内存的所有权移出函数？

Question

我正在尝试编写一个将命令行实用程序（image-magick）的标准输出加载到结构成员中的函数。我认为由于图像可以是MB，所以我最好避免复制。

/// An image data container used internally.
/// Images are 8-bit single channel for now.
pub struct Image<'a> {
    /// Width of the image in pixels.
    pub width: u32,
    /// Height of the image in pixels.
    pub height: u32,
    /// The buffer containing the image data, as a slice.
    pub pixels: &'a [u8],
}

// Load a file by first using imageMagick to convert it to a .pgm file.
fn load_using_magick<'a>(path: Path) -> Image<'a> {
    use std::io::Command;

    let output:IoResult<ProcessOutput> = Command::new("convert")
        .arg("-format pgm")
        .arg("-depth 8")
        .arg(path)
        .arg("-")
        .output();
    let output_unwrapped:ProcessOutput = match output {
        Ok(o) => o,
        Err(e) => panic!("Unable to run ImageMagick's convert tool in a separate process! convert returned: {}", e),
    };

    let bytes: &[u8] = match output_unwrapped.status.success() {
        false => panic!("signal or wrong error code from sub-process?"),
        true => output_unwrapped.output.as_slice(),
    };
    // Note, width and height will eventually get parsed out of "bytes"
    // and the returned Imaeg.pixels will point to a slice of the (already)
    // heap allocated memory.  I just need to figure out ownership first...
    Image{width:10,height:10,pixels:bytes}

}

当标准库std :: io :: Process :: Command :: output（）被调用时，我想保留的一大堆内存是由标准库（或内核？）分配的。 >

使用借阅检查器编译失败：

src/loaders.rs:41:17: 41:40 error: `output_unwrapped.output` does not live long enough
src/loaders.rs:41         true => output_unwrapped.output.as_slice(),
                                  ^~~~~~~~~~~~~~~~~~~~~~~
src/loaders.rs:21:51: 48:2 note: reference must be valid for the lifetime 'a as defined on the block at 21:50...
src/loaders.rs:21 fn load_using_magick<'a>(path: Path) -> Image<'a> {
src/loaders.rs:22     use std::io::Command;
src/loaders.rs:23 
src/loaders.rs:24     let output:IoResult<ProcessOutput> = Command::new("convert")
src/loaders.rs:25         .arg("-format pgm")
src/loaders.rs:26         .arg("-depth 8")
                  ...
src/loaders.rs:21:51: 48:2 note: ...but borrowed value is only valid for the block at 21:50
src/loaders.rs:21 fn load_using_magick<'a>(path: Path) -> Image<'a> {
src/loaders.rs:22     use std::io::Command;
src/loaders.rs:23 
src/loaders.rs:24     let output:IoResult<ProcessOutput> = Command::new("convert")
src/loaders.rs:25         .arg("-format pgm")
src/loaders.rs:26         .arg("-depth 8")
这对我来说很有意义；我试图保留的所有数据块都超出了范围，在值返回的结构中留下了一个悬空指针。那么，在返回存储区之前，我该如何将其所有权真正转移到结构中呢？ 

我已经读过Rust Lifetimes Manual。

我正在尝试编写一个将命令行实用程序（image-magick）的标准输出加载到结构成员中的函数。我认为由于图像可以是MB，所以我最好避免...

Answer 1

您的签名声称，无论呼叫者希望的生存时间如何，您都可以返回Image<'that_lifetime>。实际上，您声称要返回Image<'static>，其中包含&'static [u8]，即指向在整个程序执行过程中都有效的字节片的指针。显然，您实际从中获取字节片的ProcessOutput寿命不长。