如何在Rust中用format!宏生成无换行空格的格式化多行原始字符串？

原生方法：手动控制缩进与换行 #

如果不想引入第三方库，有两种思路解决问题：

方式1：顶格编写原始字符串 #

直接把多行字符串内容顶格书写，既保留原始字符串无需转义引号、不用行末反斜杠的优势，又不会产生多余前置空格：

fn main() {
    let name = "Alice";
    let age = 30;
    let msg = format!(r#"Hello, my name is {name}.
I'm {age} years old.
I love writing Rust code!"#);
    
    println!("{}", msg);
}

输出结果：

Hello, my name is Alice.
I'm 30 years old.
I love writing Rust code!

方式2：处理带缩进的原始字符串 #

如果为了代码结构美观必须给字符串加缩进，可以通过字符串方法去除多余空白：

fn main() {
    let name = "Bob";
    let age = 25;
    let raw_str = r#"
        Hello, {name},
        You are {age} years old.
        Glad to see you!
    "#;
    
    // 先去掉开头换行，再逐行修剪前置空格后重新拼接
    let processed_str = raw_str
        .trim_start()
        .lines()
        .map(|line| line.trim_start())
        .collect::<Vec<_>>()
        .join("\n");
    
    let msg = format!(processed_str);
    println!("{}", msg);
}

输出结果：

Hello, Bob,
You are 25 years old.
Glad to see you!

第三方库方案：使用indoc crate（推荐） #

indoc是专门处理缩进多行字符串的Rust库，能自动去除每行的共同前置缩进，同时保留原始字符串的所有优势，是最省心的方案。

步骤1：添加依赖到Cargo.toml #

[dependencies]
indoc = "2.0"

步骤2：用indoc!宏处理字符串 #

use indoc::indoc;

fn main() {
    let name = "Charlie";
    let age = 35;
    let msg = format!(indoc! {r#"
        Hello, {name},
        Your age is {age}.
        This string has proper indentation in code,
        but no extra spaces in output!
    "#});
    
    println!("{}", msg);
}

输出结果：

Hello, Charlie,
Your age is 35.
This string has proper indentation in code,
but no extra spaces in output!

内容的提问来源于stack exchange，提问作者user2138149