作为Rust的新手，我的理解是显式生命期有两个目的。

Putting an explicit lifetime annotation on a function restricts the type of code that may appear inside that function. Explicit lifetimes allow the compiler to ensure that your program is doing what you intended. If you (the compiler) want(s) to check if a piece of code is valid, you (the compiler) will not have to iteratively look inside every function called. It suffices to have a look at the annotations of functions that are directly called by that piece of code. This makes your program much easier to reason about for you (the compiler), and makes compile times managable.

第一点。，考虑以下用Python编写的程序:

import pandas as pd
import numpy as np

def second_row(ar):
    return ar[0]

def work(second):
    df = pd.DataFrame(data=second)
    df.loc[0, 0] = 1

def main():
    # .. load data ..
    ar = np.array([[0, 0], [0, 0]])

    # .. do some work on second row ..
    second = second_row(ar)
    work(second)

    # .. much later ..
    print(repr(ar))

if __name__=="__main__":
    main()

打印出来

array([[1, 0],
       [0, 0]])

这种行为总是让我吃惊。实际上，df与ar共享内存，所以当df的某些内容在work中发生变化时，这些变化也会影响ar。然而，在某些情况下，出于内存效率的原因(无复制)，这可能正是您想要的。这段代码中真正的问题是，函数second_row返回第一行而不是第二行;祝你调试好运。

考虑一个用Rust编写的类似程序:

#[derive(Debug)]
struct Array<'a, 'b>(&'a mut [i32], &'b mut [i32]);

impl<'a, 'b> Array<'a, 'b> {
    fn second_row(&mut self) -> &mut &'b mut [i32] {
        &mut self.0
    }
}

fn work(second: &mut [i32]) {
    second[0] = 1;
}

fn main() {
    // .. load data ..
    let ar1 = &mut [0, 0][..];
    let ar2 = &mut [0, 0][..];
    let mut ar = Array(ar1, ar2);

    // .. do some work on second row ..
    {
        let second = ar.second_row();
        work(second);
    }

    // .. much later ..
    println!("{:?}", ar);
}

编译这个，你得到

error[E0308]: mismatched types
 --> src/main.rs:6:13
  |
6 |             &mut self.0
  |             ^^^^^^^^^^^ lifetime mismatch
  |
  = note: expected type `&mut &'b mut [i32]`
             found type `&mut &'a mut [i32]`
note: the lifetime 'b as defined on the impl at 4:5...
 --> src/main.rs:4:5
  |
4 |     impl<'a, 'b> Array<'a, 'b> {
  |     ^^^^^^^^^^^^^^^^^^^^^^^^^^
note: ...does not necessarily outlive the lifetime 'a as defined on the impl at 4:5
 --> src/main.rs:4:5
  |
4 |     impl<'a, 'b> Array<'a, 'b> {
  |     ^^^^^^^^^^^^^^^^^^^^^^^^^^

事实上你得到了两个错误，还有一个是'a '和'b的角色互换了。查看second_row的注释，我们发现输出应该是&mut &'b mut [i32]，也就是说，输出应该是一个引用，引用的生命期为'b (Array的第二行生命期)。但是，因为我们返回的是第一行(它的生存期为'a)，编译器会报错生存期不匹配。在正确的地方。在正确的时间。调试是轻而易举的事。

作为Rust的新手，我的理解是显式生命期有两个目的。

Putting an explicit lifetime annotation on a function restricts the type of code that may appear inside that function. Explicit lifetimes allow the compiler to ensure that your program is doing what you intended. If you (the compiler) want(s) to check if a piece of code is valid, you (the compiler) will not have to iteratively look inside every function called. It suffices to have a look at the annotations of functions that are directly called by that piece of code. This makes your program much easier to reason about for you (the compiler), and makes compile times managable.

第一点。，考虑以下用Python编写的程序:

import pandas as pd
import numpy as np

def second_row(ar):
    return ar[0]

def work(second):
    df = pd.DataFrame(data=second)
    df.loc[0, 0] = 1

def main():
    # .. load data ..
    ar = np.array([[0, 0], [0, 0]])

    # .. do some work on second row ..
    second = second_row(ar)
    work(second)

    # .. much later ..
    print(repr(ar))

if __name__=="__main__":
    main()

打印出来

array([[1, 0],
       [0, 0]])

这种行为总是让我吃惊。实际上，df与ar共享内存，所以当df的某些内容在work中发生变化时，这些变化也会影响ar。然而，在某些情况下，出于内存效率的原因(无复制)，这可能正是您想要的。这段代码中真正的问题是，函数second_row返回第一行而不是第二行;祝你调试好运。

考虑一个用Rust编写的类似程序:

#[derive(Debug)]
struct Array<'a, 'b>(&'a mut [i32], &'b mut [i32]);

impl<'a, 'b> Array<'a, 'b> {
    fn second_row(&mut self) -> &mut &'b mut [i32] {
        &mut self.0
    }
}

fn work(second: &mut [i32]) {
    second[0] = 1;
}

fn main() {
    // .. load data ..
    let ar1 = &mut [0, 0][..];
    let ar2 = &mut [0, 0][..];
    let mut ar = Array(ar1, ar2);

    // .. do some work on second row ..
    {
        let second = ar.second_row();
        work(second);
    }

    // .. much later ..
    println!("{:?}", ar);
}

编译这个，你得到

error[E0308]: mismatched types
 --> src/main.rs:6:13
  |
6 |             &mut self.0
  |             ^^^^^^^^^^^ lifetime mismatch
  |
  = note: expected type `&mut &'b mut [i32]`
             found type `&mut &'a mut [i32]`
note: the lifetime 'b as defined on the impl at 4:5...
 --> src/main.rs:4:5
  |
4 |     impl<'a, 'b> Array<'a, 'b> {
  |     ^^^^^^^^^^^^^^^^^^^^^^^^^^
note: ...does not necessarily outlive the lifetime 'a as defined on the impl at 4:5
 --> src/main.rs:4:5
  |
4 |     impl<'a, 'b> Array<'a, 'b> {
  |     ^^^^^^^^^^^^^^^^^^^^^^^^^^

事实上你得到了两个错误，还有一个是'a '和'b的角色互换了。查看second_row的注释，我们发现输出应该是&mut &'b mut [i32]，也就是说，输出应该是一个引用，引用的生命期为'b (Array的第二行生命期)。但是，因为我们返回的是第一行(它的生存期为'a)，编译器会报错生存期不匹配。在正确的地方。在正确的时间。调试是轻而易举的事。

If a function receives two references as arguments and returns a reference, then the implementation of the function might sometimes return the first reference and sometimes the second one. It is impossible to predict which reference will be returned for a given call. In this case, it is impossible to infer a lifetime for the returned reference, since each argument reference may refer to a different variable binding with a different lifetime. Explicit lifetimes help to avoid or clarify such a situation.

同样地，如果一个结构体包含两个引用(作为两个成员字段)，那么该结构体的成员函数可能有时返回第一个引用，有时返回第二个引用。同样，显式的生存期可以防止这种歧义。

在一些简单的情况下，存在生命期省略，编译器可以推断生命期。

我在这里找到了另一个很好的解释:http://doc.rust-lang.org/0.12.0/guide-lifetimes.html#returning-references。

一般情况下，只有当引用是时才有可能返回 从过程的参数派生。在这种情况下，是指针 Result将始终具有与其中一个参数相同的生命周期; 命名的生命周期指示是哪个参数。

你的例子不能工作的原因很简单，因为Rust只有局部生存期和类型推断。你的建议需要全局推理。当您有一个生命周期不能被省略的引用时，必须对其进行注释。

以下结构中的生命周期注释:

struct Foo<'a> {
    x: &'a i32,
}

指定一个Foo实例不应该比它包含的引用活得更久(x字段)。

你在Rust书中看到的例子并没有说明这一点，因为f和y变量同时超出范围。

一个更好的例子是:

fn main() {
    let f : Foo;
    {
        let n = 5;  // variable that is invalid outside this block
        let y = &n;
        f = Foo { x: y };
    };
    println!("{}", f.x);
}

f确实比f。x指向的变量寿命长。