这个 Rust 例子里面怎么对字符串赋值

› github.com/mozilla/rust

This topic created in 1802 days ago, the information mentioned may be changed or developed.

利用 match 循环匹配数组中的值，如果是 Apple，改成 RedApple：

let mut names = ["Apple", "Banana"];
for name in names.iter_mut() {
	*name = match name {
    	&mut "Apple" => { format!("Red{}", name).as_str() }
        _ => "Hello"
    }
}

报错：

&mut "Apple" => { format!("Red{}", name).as_str() }
- temporary value is freed at the end of this statement
    | |                               |
    | |                               creates a temporary which is freed while still in use
    _________- borrow later used here
    |
    = note: consider using a `let` binding to create a longer lived value

8 replies • 2021-06-06 18:12:09 +08:00

shidenggui

Jun 6, 2021

let mut names: Vec<String> = ["Apple", "Banana"].iter().map(|s| s.to_string()).collect();
for name in names.iter_mut() {
*name = match name.as_str() {
"Apple" => { format!("Red{}", name) }
_ => "Hello".to_string()
}
}

Jirajine

Jun 6, 2021

因为你这个数组里面存的是&str，也就是指向字符串的指针。你要把它改成别的，就得让它指向其他的字符串。而你代码里只创建了个临时的字符串，没有绑定到变量。
```rust
let mut names = ["Apple", "Banana"];
let red_apple = format!("Red{}", name);
for name in names.iter_mut() {
*name = match name {
&mut "Apple" => { red_apple.as_str() }
_ => "Hello"
}
}

```

longkas239

Jun 6, 2021

@shidenggui @Jirajine 主要是想动态的组装一个字符串赋值过去，所以不把数组中的类型换成 String 就没有办法这么做是吧。能把堆上存储的 String 复制到栈上变成&str 吗

shidenggui

Jun 6, 2021

@longkas239 可以啊，只要你能接受对应的字符串内存泄漏就行

let mut names = ["Apple", "Banana"];
for name in names.iter_mut() {
*name = match name {
&mut "Apple" => { Box::leak(Box::new(format!("Red{}", name)) )}
_ => "Hello"
}
}

Jirajine

Jun 6, 2021

@longkas239 #3 当然可以，像我写的那样不就是实现了动态组装一个字符串，然后把引用存到栈上么。并且你的 string literal 是存到静态资源里，也不是栈上，标准库里的 String 类型不能存到栈上，栈上只能存引用。

当然如果你想避免堆分配，可以看一下 https://lib.rs/crates/inlinable_string

12101111

Jun 6, 2021

你的数组里存的是&'static str, 即只读的字符串, 你的 names 应该是 String 的数组

VDimos

Jun 6, 2021 via Android

names 是在栈里面的，你 format 的变量离开作用域就释放了，得用一些手段避开这些操作。
fn main() {
let mut names = ["A", "B"];
for name in names.iter_mut() {
*name = match name {
&mut "A" => {
let a = format!("GGG{}", name);
let r = unsafe {
let p = a.as_str() as *const str;
let s = &*p as &str;
s
};
std::mem::forget(a);
r
},
_ => "hello"
}
}

println!("{:#?}", names);
}

PeterD

Jun 6, 2021

names 是 [&'static str]。format!("Red{}", name) 不是 'static

```rust
pub fn main() {
let mut names = ["Apple".to_string(), "Banana".to_string()];
for name in names.iter_mut() {
match name.as_str() {
"Apple" => *name = format!("Red{}", name),
_ => *name = "Hello".to_string(),
}
}
}
```