问题描述
考虑此函数,该函数应返回给定Path
的文件扩展名.
Consider this function that should return the file extension of a given Path
.
pub fn get_extension<'a>(path: &'a Path) -> Option<&'a str> {
let path_str = path.as_str().unwrap();
let ext_pos = regex!(".[a-z0-9]+$").find(path_str);
match ext_pos {
Some((start, _)) => {
return Some(path_str.as_slice().slice_from(start))
},
None => return None
}
}
错误消息如下:
`path_str` does not live long enough
错误消息非常清楚,很遗憾我不能自己解决.我从理论上理解它,但对我来说仍然有一些模糊的东西.
The error message is pretty clear and it's a shame I can't work it out on my own. I understand it in theory but there are still a couple of blurred things for me.
我了解编译器想告诉我path_str
的生存期不足以使其有效,因为带有生命周期'a
的返回值标记为.
I understand that the compiler wants to tell me that path_str
does not live long enough to be valid as the return value with is marked with lifetime 'a
.
但是这对我而言止于此:
But this is where it stops for me:
-
我了解对
path
的引用(输入参数)的寿命应与对包裹在Option
的str
引用(输出参数)的寿命完全一样长
I understand that the reference to
path
(the input parameter) should life exactly as long as the reference to thestr
that is wrapped in theOption
(the output parameter)
因为我们返回了Some(path_str.as_slice().slice_from(start))
,所以我认为在实践中这意味着path_str
的寿命必须与path
一样长.
since we return Some(path_str.as_slice().slice_from(start))
I assume that in practice that means that path_str
needs to live as long as path
.
我不明白的是为什么准确?path_str
的寿命不够长,我该如何解决?是什么让它很快死掉?
What I don't understand is why exactly does path_str
not live long enough and how could I fix this? What makes it die to soon?
更新
删除多余的as_slice()
可使代码编译.有人知道为什么吗?还指出,存在方法直接获取扩展名.但是,是的,尽管如此,我实际上对学习问题背后的故事更感兴趣.
As pointed out in the comments and also on IRC removing the superflous as_slice()
makes the code compile. Does anyone know why that is? It was also pointed out that there exists a method to get the extension directly. But yep, I'm actually more interested in learning the story behind the problem though.
推荐答案
这不是错误.这里的问题"是 as_slice
的定义.它接受对其参数的引用,并返回一个与 reference 相同的生存期的&str
,它不能自省其被调用的任何类型的内部生存期.也就是说,path_str.as_slice()
返回的&str
持续时间与path_str
相同,而不是,只要数据path_str
指向(原始Path
).
This isn't a bug. The "problem" here is as_slice
's definition. It takes a reference to its arguments, and returns a &str
with the same lifetime as the reference, it can't introspect into the internal lifetimes of whatever type it is being called on. That is, path_str.as_slice()
returns a &str
that lasts for as long as path_str
, not as long as the data path_str
points at (the original Path
).
换句话说,这里有两个生命.我将在@Arjan的提交的错误的示例中使用假设的块寿命注释语法基于我的回复).
In other words, there's two lifetimes here. I'll use a hypothetical block-lifetime annotation syntax on the example from @Arjan's filed bug (this answer is based of my response there).
fn test<'a>(s: &'a String) -> &'a str {
'b: {
let slice: &'a str = s.as_slice();
slice.as_slice()
}
}
对于第二个as_slice
调用,我们有self: &'b &'a str
,因此它返回&'b str
,这太短了:'b
只是test
的本地.
For the second as_slice
call we have self: &'b &'a str
, and thus it returns &'b str
, which is too short: 'b
is just local to test
.
正如您所发现的,现在的解决方法是删除无关的as_slice
调用.但是,使用动态大小类型(DST),我们将能够写impl StrSlice for str
,然后slice.as_slice()
将返回&'a str
,因为不会再增加引用层(即self: &'a str
).
As you discovered, the fix now is just removing the extraneous as_slice
call. However, with dynamically sized types (DST), we will be able to write impl StrSlice for str
, and then slice.as_slice()
will be returning a &'a str
, since there won't be an extra layer of references (that is, self: &'a str
).
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