如何匹配从标准输入读取的字符串? [英] How do I match on a string read from standard input?

问题描述

在学习 Rust 的练习中，我正在尝试一个简单的程序，该程序将接受您的姓名，然后在有效时打印您的姓名.

In an exercise to learn Rust, I'm trying a simple program that will accept your name, then print your name if it's Valid.

只有Alice"和Bob"是有效的名字.

use std::io;

fn main() {
    println!("What's your name?");
    let mut name = String::new();

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

    greet(&name);
}

fn greet(name: &str) {
    match name {
        "Alice" => println!("Your name is Alice"),
        "Bob"   => println!("Your name is Bob"),
        _ => println!("Invalid name: {}", name),
    }
}

当我cargo run这个main.rs文件时，我得到:

When I cargo run this main.rs file, I get:

What's your name?
Alice
Invalid name: Alice

现在，我的猜测是，因为Alice"的类型是 &'static str 而 name 的类型是 &str, 可能是匹配不正确...

Now, my guess is, because "Alice" is of type &'static str and name is of type &str, maybe it's not matching correctly...

推荐答案

我敢打赌这不是由类型不匹配引起的.我打赌有一些不可见的字符(在这种情况下是新行).为了实现您的目标，您应该修剪输入字符串:

I bet that it isn't caused by type mismatch. I place my bet on that there are some invisible characters (new line in this case). To achieve your goal you should trim your input string:

match name.trim() {
    "Alice" => println!("Your name is Alice"),
    "Bob"   => println!("Your name is Bob"),
    _ => println!("Invalid name: {}", name),
}

这篇关于如何匹配从标准输入读取的字符串?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！