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"如何分析与方差相关的通用生命周期"

[英]How to analyze generic lifetime when it is relevant to variance

原文 2022-02-06 03:47:12 3 1 rust/ lifetime/ variance/ invariance

 0 // code snippet 1 1 2 struct MutStr<'a >{ 3 s: &'a mut &'a str, 4 } 5 6 fn main() { 7 let mut s: &'static str = "hello"; 8 *MutStr{ 9 s: &mut s, 10 }.s = "world"; 11 println!("{}", s); 12 }

1 个解决方案

I have read this article, knowing some basic concepts about variance, such as covariance invariance and contravariance.我读过这篇文章，了解了一些关于方差的基本概念，例如协方差不变性和逆变性。 And I kind of think my question relate to that, but don't know how to use that to analyze code snippet 1.而且我有点认为我的问题与此有关，但不知道如何使用它来分析代码片段 1。
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You are on the right track, the difference does lie with lifetime variance.您走在正确的轨道上，差异确实在于终生差异。 There is a table in the Rust Reference 10.5 Subtyping and Variance<\/a> that I think is helpful: Rust Reference 10.5 Subtyping and Variance<\/a>中有一张我认为很有帮助的表格：

Type类型<\/th> Variance in 'a<\/code> 'a<\/code>方差<\/th> Variance in T<\/code> T<\/code>的方差<\/th><\/tr><\/thead>

&'a T<\/code><\/td> covariant协变<\/td> covariant协变<\/td><\/tr>

&'a mut T<\/code><\/td> covariant协变<\/td> invariant不变的<\/td><\/tr><\/tbody><\/table>
In your second snippet, your references are immutable meaning the lifetime associated with them can be shortened as necessary.在您的第二个片段中，您的引用是不可变的，这意味着可以根据需要缩短与它们相关的生命周期。 A reference to the variable y<\/code> cannot lengthen its lifetime so the reference to x<\/code> must be shortened.对变量y<\/code>的引用不能延长其生命周期，因此必须缩短对x<\/code>的引用。 An thus the reference bound to r<\/code> is tied to the lifetime of y<\/code> and therefore you get an error when you try to use r<\/code> after y<\/code> has gone out of scope.因此，绑定到r<\/code>的引用与y<\/code>的生命周期相关联，因此在y<\/code>超出范围后尝试使用r<\/code>时会出现错误。
In the first snippet however, you have a mutable reference to a &'a str<\/code> .然而，在第一个片段中，您有一个对&'a str<\/code>的可变引用。 If you look at the table above, you'll see that types referenced by a mutable reference are invariant and since the type is itself a &'a str<\/code> , that means that 'a<\/code> is invariant.如果您查看上表，您会发现可变引用所引用的类型是不变的，并且由于该类型本身是&'a str<\/code> ，这意味着'a<\/code>是不变的。 This means, unlike in the second snippet, the compiler can not<\/em> shorten the lifetime of 'a<\/code> at all.这意味着，与第二个片段不同，编译器根本无法<\/em>缩短'a<\/code>的生命周期。 So when you try to use s<\/code> to make a MutStr<\/code> , the compiler sees that you're passing a &'static str<\/code> that it cannot shorten, so 'a<\/code> must be 'static<\/code> .因此，当您尝试使用s<\/code>制作MutStr<\/code>时，编译器会看到您正在传递一个无法缩短的&'static str<\/code> ，因此'a<\/code>必须是'static<\/code> 。 But then it tries to reconcile that 'a<\/code> is also linked to the variable s<\/code> which is not 'static<\/code> , so you get the error.但随后它试图调和'a<\/code>也链接到不是'static<\/code>的变量s<\/code> ，所以你得到了错误。
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Type类型<\/th>	Variance in `'a<\/code>` `'a<\/code>方差<\/th>`	Variance in `T<\/code>` `T<\/code>的方差<\/th><\/tr><\/thead>`
`&'a T<\/code><\/td>`	covariant协变<\/td>	covariant协变<\/td><\/tr>
`&'a mut T<\/code><\/td>`	covariant协变<\/td>	invariant不变的<\/td><\/tr><\/tbody><\/table> In your second snippet, your references are immutable meaning the lifetime associated with them can be shortened as necessary.在您的第二个片段中，您的引用是不可变的，这意味着可以根据需要缩短与它们相关的生命周期。 A reference to the variable `y<\/code> cannot lengthen its lifetime so the reference to x<\/code> must be shortened.``对变量y<\/code>的引用不能延长其生命周期，因此必须缩短对x<\/code>的引用。 An thus the reference bound to r<\/code> is tied to the lifetime of y<\/code> and therefore you get an error when you try to use r<\/code> after y<\/code> has gone out of scope.因此，绑定到r<\/code>的引用与y<\/code>的生命周期相关联，因此在y<\/code>超出范围后尝试使用r<\/code>时会出现错误。` In the first snippet however, you have a mutable reference to a &'a str<\/code> .然而，在第一个片段中，您有一个对&'a str<\/code>的可变引用。 If you look at the table above, you'll see that types referenced by a mutable reference are invariant and since the type is itself a &'a str<\/code> , that means that 'a<\/code> is invariant.如果您查看上表，您会发现可变引用所引用的类型是不变的，并且由于该类型本身是&'a str<\/code> ，这意味着'a<\/code>是不变的。 This means, unlike in the second snippet, the compiler can not<\/em> shorten the lifetime of 'a<\/code> at all.这意味着，与第二个片段不同，编译器根本无法<\/em>缩短'a<\/code>的生命周期。 So when you try to use s<\/code> to make a MutStr<\/code> , the compiler sees that you're passing a &'static str<\/code> that it cannot shorten, so 'a<\/code> must be 'static<\/code> .因此，当您尝试使用s<\/code>制作MutStr<\/code>时，编译器会看到您正在传递一个无法缩短的&'static str<\/code> ，因此'a<\/code>必须是'static<\/code> 。 But then it tries to reconcile that 'a<\/code> is also linked to the variable s<\/code> which is not 'static<\/code> , so you get the error.但随后它试图调和'a<\/code>也链接到不是'static<\/code>的变量s<\/code> ，所以你得到了错误。 *`"`*