打印从stdin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?
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问题描述
我试图编写一些代码,该代码从stdin读取名称并打印出来.问题是在打印变量后,该行立即中断,并且该变量之后的字符在下一行中打印:
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);
}
'!'在下一行而不是预期位置打印.
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.
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()
只能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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