为什么不可变字符串可以调用 String::add(mut self, other: &str) [英] Why immutable string can call String::add(mut self, other: &str)

问题描述

在stdlibstring.rs:

In stdlib string.rs:

impl Add<&str> for String {
    type Output = String;

    #[inline]
    fn add(mut self, other: &str) -> String {
        self.push_str(other);
        self
    }
}


let s1 = String::from("tic");
let s2 = String::from("tac");

let s = s1 + &s2;// it works

s1 在这里是不可变的，但是 Add::add(mut self, other: &str) 是 mut，我只是想知道为什么.

s1 is immutable here, but Add::add(mut self, other: &str) is mut, I just want to know why.

推荐答案

当您在串联中使用 s1 时，您可以借用和使用它.如果您尝试在该行之后打印 s1let s = s1 + &s2;//有效,会报错，因为是move后用的:

You borrow and consume s1 when you use it in the concatenation. If you try to print s1 after the line let s = s1 + &s2;// it works, there will be an error because it is used after move:

3 |     let s1 = String::from("tic");
  |         -- move occurs because `s1` has type `std::string::String`, which does not implement the `Copy` trait
...
6 |     let s = s1 + &s2;// it works
  |             -- value moved here
7 |     println!("{}{}{}",s,s1,s2);
  |                         ^^ value borrowed here after move

s1 不需要是可变的，因为变量永远不会变异，数据被移动并变异到变量 s.

s1 does not need to be mutable as the variable is never mutated the data is moved and mutated to the variable s.

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