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# 生命周期过大-02
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继上篇文章后，我们再来看一段**可能**涉及生命周期过大导致的无法编译问题:
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```rust
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fn bar(writer: &mut Writer) {
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baz(writer.indent());
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writer.write("world");
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}
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fn baz(writer: &mut Writer) {
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writer.write("hello");
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}
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pub struct Writer<'a> {
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target: &'a mut String,
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indent: usize,
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}
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impl<'a> Writer<'a> {
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fn indent(&'a mut self) -> &'a mut Self {
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&mut Self {
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target: self.target,
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indent: self.indent + 1,
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}
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}
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fn write(&mut self, s: &str) {
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for _ in 0..self.indent {
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self.target.push(' ');
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}
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self.target.push_str(s);
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self.target.push('\n');
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}
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}
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fn main() {}
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```
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报错如下：
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```console
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error[E0623]: lifetime mismatch
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--> src/main.rs:2:16
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1 | fn bar(writer: &mut Writer) {
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| -----------
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| |
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| these two types are declared with different lifetimes...
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2 | baz(writer.indent());
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| ^^^^^^ ...but data from `writer` flows into `writer` here
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```
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WTF，这什么报错，之前都没有见过，而且很难理解，什么叫`writer`滑入了另一个`writer`？
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别急，我们先来仔细看下代码，注意这一段：
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```rust
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impl<'a> Writer<'a> {
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fn indent(&'a mut self) -> &'a mut Self {
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&mut Self {
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target: self.target,
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indent: self.indent + 1,
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}
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}
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```
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这里的生命周期定义说明`indent`方法使用的。。。等等！你的代码错了，你怎么能在一个函数中返回一个新创建实例的引用？！！最重要的是，编译器不提示这个错误，竟然提示一个莫名其妙看不懂的东东。
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行，那我们先解决这个问题，将该方法修改为:
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```rust
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fn indent(&'a mut self) -> Writer<'a> {
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Writer {
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target: self.target,
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indent: self.indent + 1,
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}
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}
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```
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怀着惴惴这心，再一次运行程序，果不其然，编译器又朝我们扔了一坨错误：
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```console
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error[E0308]: mismatched types
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--> src/main.rs:2:9
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2 | baz(writer.indent());
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| ^^^^^^^^^^^^^^^
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| |
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| expected `&mut Writer<'_>`, found struct `Writer`
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| help: consider mutably borrowing here: `&mut writer.indent()`
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```
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哦，这次错误很明显，因为`baz`需要`&mut Writer`，但是咱们`writer.indent`返回了一个`Writer`，因此修改下即可:
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```rust
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fn bar(writer: &mut Writer) {
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baz(&mut writer.indent());
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writer.write("world");
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}
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```
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这次总该成功了吧？再次心慌慌的运行编译器，哐：
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```console
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error[E0623]: lifetime mismatch
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--> src/main.rs:2:21
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1 | fn bar(writer: &mut Writer) {
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| -----------
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| |
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| these two types are declared with different lifetimes...
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2 | baz(&mut writer.indent());
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| ^^^^^^ ...but data from `writer` flows into `writer` here
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```
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可恶，还是这个看不懂的错误，仔细检查了下代码，这次真的没有其他错误了，只能硬着头皮上。
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大概的意思可以分析，生命周期范围不匹配，说明一个大一个小，然后一个`writer`中流入到另一个`writer`说明，两个`writer`的生命周期定义错了，既然这里提到了`indent`方法调用，那么我们再去仔细看一眼：
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```rust
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impl<'a> Writer<'a> {
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fn indent(&'a mut self) -> Writer<'a> {
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Writer {
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target: self.target,
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indent: self.indent + 1,
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}
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}
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...
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}
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```
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好像有点问题，`indent`返回的`Writer`的生命周期和外面调用者的`Writer`的生命周期一模一样，这很不合理，一眼就能看出前者远小于后者。
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这里稍微展开以下，为何`indent`方法返回值的生命周期不能与参数中的`self`相同。**首先，我们假设它们可以相同，也就是上面的代码可以编译通过**，由于此时在返回值中借用了`self`的可变引用，意味着**如果你在返回值被使用后，还继续使用`self` 会导致重复借用的错误，因为返回值的生命周期将持续到 `self` 结束**。
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既然不能相同，那我们尝试着修改下`indent`:
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```rust
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fn indent<'b>(&'b mut self) -> Writer<'b> {
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Writer {
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target: self.target,
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indent: self.indent + 1,
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}
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}
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```
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Bang! 编译成功，不过稍等，回想下生命周期消除的规则，我们还可以实现的更优雅：
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```rust
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fn bar(writer: &mut Writer) {
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baz(&mut writer.indent());
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writer.write("world");
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}
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fn baz(writer: &mut Writer) {
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writer.write("hello");
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}
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pub struct Writer<'a> {
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target: &'a mut String,
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indent: usize,
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}
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impl<'a> Writer<'a> {
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fn indent(&mut self) -> Writer {
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Writer {
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target: self.target,
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indent: self.indent + 1,
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}
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}
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fn write(&mut self, s: &str) {
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for _ in 0..self.indent {
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self.target.push(' ');
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}
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self.target.push_str(s);
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self.target.push('\n');
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}
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}
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fn main() {}
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```
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至此，问题彻底解决，太好了，我感觉我又变强了。可是默默看了眼自己的头发，只能以`哎～`一声叹息结束本章内容。 |
