Question

我在做Rust Koans 我陷入了一个问题：

#[test]
fn for_loops_two() {
    let words: [&'static str; 3] = ["I", "love", "Rust"];
    let space: &str = " ";
    let mut sentence: String = String::new();
    for word in words.iter() {
        // __
    }

    println!("{:?}", sentence);
    assert!(sentence == "I love Rust".to_string());
}

我知道我需要连接字符串，但这会失败：

#[test]
fn for_loops_two() {
    let words: [&'static str; 3] = ["I", "love", "Rust"];
    let mut sentence: String = String::new();
    for word in words.iter() {
        sentence.push_str(word);
    }

    println!("{:?}", sentence); // "ILoveRust"
    assert!(sentence == "I love Rust".to_string());
}

我可以在每次迭代后添加一个空格：

#[test]
fn for_loops_two() {
    let words: [&'static str; 3] = ["I", "love", "Rust"];
    let space: &str = " ";
    let mut sentence: String = String::new();
    for word in words.iter() {
        sentence.push_str(word);
        sentence.push_str(space);
    }

    println!("{:?}", sentence); // "I Love Rust "
    assert!(sentence == "I love Rust".to_string());
}

这也将失败，因为最后的迭代将添加一个空格。

我想如果我们在上一次迭代中，我可以写一个条件语句，但是我在努力使语法正确。而且，我觉得所有这些都有更好的解决方案，我只是想不出语法。

如何使上面的断言在循环中带有条件，而不在上一次迭代中添加空间？

Answer 1

有点晚了，但这里有一个只修改循环内块的解决方案：

#[test]
fn for_loops_two() {
    let words: [&'static str; 3] = ["I", "love", "Rust"];
    let mut sentence: String = String::new();
    for word in words.iter() {
        if sentence != "".to_string() {
            sentence.push(' ')
        }
        sentence.push_str(word)
    }
    println!("{:?}", sentence);
    assert!(sentence == "I love Rust".to_string());
}

Answer 2

您可以使用slice::join：

#[test]
fn for_loops_two() {
    let words: [&'static str; 3] = ["I", "love", "Rust"];

    let sentence = words.join(" ");

    assert!(sentence == "I love Rust".to_string());
}

关于链接的SliceConcatExt特性的注释：在文档中被列为不稳定，但是方法是稳定的-在当前的稳定版Rust上，上述编译就很好。 / p>

如果您希望遵守koan的约束条件并使用for循环，则可以按照您的建议使用if（使用{{3}来确定您是否最后}），或enumerate字符串末尾的最后一个空格：

#[test]
fn for_loops_two_with_len_check() {
    let words: [&'static str; 3] = ["I", "love", "Rust"];
    const SPACE: char = ' ';
    let number_of_words = words.len();
    let mut sentence = String::new();

    for (i, word) in words.iter().enumerate() {
        sentence.push_str(word);
        if i < number_of_words-1 {
            sentence.push(SPACE);
        }
    }

    assert!(sentence == "I love Rust".to_string());
}


#[test]
fn for_loops_two_with_pop() {
    let words: [&'static str; 3] = ["I", "love", "Rust"];
    const SPACE: char = ' ';
    let mut sentence = String::new();

    for word in words.iter() {
        sentence.push_str(word);
        sentence.push(SPACE);
    }
    let _ = sentence.pop();

    assert!(sentence == "I love Rust".to_string());
}

仅使用for循环体内的条件将带空格的字符串连接起来

2 个答案: