如何在不接收“未使用变量"的情况下获得用户输入?警告? [英] How can I get user input without receiving an "Unsed Variable" warning?

问题描述

我正在研究Rust，并决定构建一个小程序，该程序可以接受用户的输入并进行打印，但是还希望对其进行一些数学上的练习.目前，这是我接受用户输入的方式:

I'm taking a look at Rust and decided to build a small program that takes a user's input and prints it, but also want to do some math stuff with it for practice. Currently, this is how I am taking user input:

let mut number = String::new();

let input = io::stdin().read_line(&mut number)
    .ok()
    .expect("Failed to read line");

println!("You entered {}", number);

但是，尽管我确实以这种方式获得了正确的输入，但Cargo还是给了我以下警告:

However, although I do get the correct input this way, Cargo gives me the following warning:

src/main.rs:10:9:10:14警告:未使用的变量: input ，默认情况下#[warn(unused_variables)]启用

src/main.rs:10:9: 10:14 warning: unused variable: input, #[warn(unused_variables)] on by default

src/main.rs:10让input = reader.read_line(& mut number)

src/main.rs:10 let input = reader.read_line(&mut number)

如果我只使用 input 变量，无论我输入什么数字，我都会得到一个"2".作为回报，当我打印号码时.

If I were to just use the input variable, no matter what number I enter I would get a "2" in return when I print the number.

如何避免该警告?是否有另一种方法可以在不创建2个变量绑定的情况下接受输入?

How can I avoid the warning? Is there another way for me to take input without creating 2 variable bindings?

推荐答案

您不能简单地将值写入变量.只要该值的类型未标记为 must_use ，您就可以忽略该值.

You can simply not write the value to a variable. As long as the type of the value is not marked must_use, you can ignore the value.

let mut number = String::new();

io::stdin().read_line(&mut number)
           .ok()
           .expect("Failed to read line");

println!("You entered {}", number);

---

[商业]

您可以将 text_io 板条箱用于

You can use the text_io crate for super short and readable input like

let i: i32 = read!()
let tup: (i32, String) = read!("{}, {}");

[/commercial]

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