打印从标准输入读取的字符串时如何忽略换行符? [英] How to ignore the line break while printing a string read from stdin?

问题描述

我尝试编写一些代码，从标准输入读取名称并打印出来.问题是打印变量后立即换行，变量后面的字符打印在下一行:

I tried to write a bit of code which reads a name from stdin and prints it. The problem is the line breaks immediately after printing the variable and the characters following the variable are printed in the next line:

use std::io;

fn main() {
    println!("Enter your name:");
    let mut name = String::new();
    io::stdin().read_line(&mut name).expect("Failed To read Input");
    println!("Hello '{}'!", name);
}

！"打印在下一行，这不是预期的位置.

The '!' is printed in the next line, which is not the expected location.

推荐答案

使用 .trim() 删除字符串上的空格.这个例子应该可以工作.

Use .trim() to remove whitespace on a string. This example should work.

use std::io;

fn main() {
    println!("Enter your name:");
    let mut name = String::new();
    io::stdin().read_line(&mut name).expect("Failed To read Input");
    println!("Hello '{}'!", name.trim());
}

还有 trim_start() 和 .trim_end() 如果您只需要从字符串的一侧删除空格更改.

There also trim_start() and .trim_end() if you need to remove whitespace changes from only one side of the string.

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