如何在Rust的while循环中将用户输入添加至Vector？

错误原因 #

报错核心是所有权转移：String类型未实现Copy trait，执行words.push(WordInput)时，WordInput的所有权会完全转移到Vec中，原变量变为无效状态。下一次循环调用io::stdin().read_line(&mut WordInput)时，试图借用已失去所有权的变量，因此触发错误。

修复方案 #

方案1：每次循环新建WordInput

将WordInput的声明移至while循环内部，每次迭代创建新的String，确保每次push的都是独立变量，不会影响下一次循环使用：

use std::io;

fn CarryOn()->i8{
    println!("Do you want to continue adding words to the glossary?");
    println!("Press 1 for yes, enter any other number for no.");
    let mut s1 = String::new();
    io::stdin().read_line(&mut s1).expect("Failed");
    let n1:i8 = s1.trim().parse().expect("Not a valid Number");
    return n1;
}

fn main(){
    let mut words: Vec<String> = Vec::new();
    println!("What's your name?");
    let mut name = String::new();
    io::stdin().read_line(&mut name).expect("Failed");
    println!("Welcome {}",name);
    let mut LastWord = 1;
    words.push("Hello".to_string());
    while LastWord == 1 {
        let mut WordInput = String::new(); // 移到循环内，每次新建变量
        println!("Please type a word to add to the glossary");
        io::stdin().read_line(&mut WordInput).expect("Failed");
        WordInput.make_ascii_lowercase();
        println!("{}",WordInput);
        words.push(WordInput);
        LastWord = CarryOn();
    }
}

方案2：使用take()方法复用变量

String::take()会拿走原字符串的所有内容，留下空的String，原变量仍保持有效，可在下一次循环中继续写入：

use std::io;

fn CarryOn()->i8{
    println!("Do you want to continue adding words to the glossary?");
    println!("Press 1 for yes, enter any other number for no.");
    let mut s1 = String::new();
    io::stdin().read_line(&mut s1).expect("Failed");
    let n1:i8 = s1.trim().parse().expect("Not a valid Number");
    return n1;
}

fn main(){
    let mut words: Vec<String> = Vec::new();
    println!("What's your name?");
    let mut name = String::new();
    io::stdin().read_line(&mut name).expect("Failed");
    println!("Welcome {}",name);
    let mut LastWord = 1;
    let mut WordInput = String::new();
    words.push("Hello".to_string());
    while LastWord == 1 {
        println!("Please type a word to add to the glossary");
        io::stdin().read_line(&mut WordInput).expect("Failed");
        WordInput.make_ascii_lowercase();
        println!("{}",WordInput);
        words.push(WordInput.take()); // 拿走内容，原变量留空可复用
        LastWord = CarryOn();
    }
}

方案对比 #

• 方案1逻辑直观，适合新手理解，每次创建新变量的性能开销可忽略。
• 方案2复用变量，避免重复内存分配，性能略优，但需要理解take()的所有权语义。

内容的提问来源于stack exchange，提问作者John