sun*_*ica 13 integer user-input input rust
我发现的现有答案都是基于from_str(例如 有效地从控制台读取用户输入),但显然from_str(x)已经改变为x.parse()Rust 1.0.作为一个新手,考虑到这一变化,原始解决方案应该如何适应并不明显.
从Rust 1.0开始,从用户那里获取整数输入的最简单方法是什么?
Mic*_*ael 33
这是一个包含所有可选类型注释和错误处理的版本,对于像我这样的初学者可能会有用:
use std::io;
fn main() {
let mut input_text = String::new();
io::stdin()
.read_line(&mut input_text)
.expect("failed to read from stdin");
let trimmed = input_text.trim();
match trimmed.parse::<u32>() {
Ok(i) => println!("your integer input: {}", i),
Err(..) => println!("this was not an integer: {}", trimmed),
};
}
Run Code Online (Sandbox Code Playgroud)
cod*_*101 11
如果您正在寻找一种读取输入的方法,以便在您无法访问的 codeforces 等网站上进行竞争性编程,那么text_io此解决方案适合您。
我使用以下宏从 读取不同的值stdin:
use std::io;
#[allow(unused_macros)]
macro_rules! read {
($out:ident as $type:ty) => {
let mut inner = String::new();
io::stdin().read_line(&mut inner).expect("A String");
let $out = inner.trim().parse::<$type>().expect("Parseble");
};
}
#[allow(unused_macros)]
macro_rules! read_str {
($out:ident) => {
let mut inner = String::new();
io::stdin().read_line(&mut inner).expect("A String");
let $out = inner.trim();
};
}
#[allow(unused_macros)]
macro_rules! read_vec {
($out:ident as $type:ty) => {
let mut inner = String::new();
io::stdin().read_line(&mut inner).unwrap();
let $out = inner
.trim()
.split_whitespace()
.map(|s| s.parse::<$type>().unwrap())
.collect::<Vec<$type>>();
};
}
Run Code Online (Sandbox Code Playgroud)
如下使用
fn main(){
read!(x as u32);
read!(y as f64);
read!(z as char);
println!("{} {} {}", x, y, z);
read_vec!(v as u32); // Reads space separated integers and stops when newline is encountered.
println!("{:?}", v);
}
Run Code Online (Sandbox Code Playgroud)
可能最简单的部分是使用text_io crate并写:
#[macro_use]
extern crate text_io;
fn main() {
// read until a whitespace and try to convert what was read into an i32
let i: i32 = read!();
println!("Read in: {}", i);
}
Run Code Online (Sandbox Code Playgroud)
如果您需要同时读取多个值,则可能需要每晚使用Rust.
以下是几种可能性(Rust 1.7):
use std::io;
fn main() {
let mut n = String::new();
io::stdin()
.read_line(&mut n)
.expect("failed to read input.");
let n: i32 = n.trim().parse().expect("invalid input");
println!("{:?}", n);
let mut n = String::new();
io::stdin()
.read_line(&mut n)
.expect("failed to read input.");
let n = n.trim().parse::<i32>().expect("invalid input");
println!("{:?}", n);
let mut n = String::new();
io::stdin()
.read_line(&mut n)
.expect("failed to read input.");
if let Ok(n) = n.trim().parse::<i32>() {
println!("{:?}", n);
}
}
Run Code Online (Sandbox Code Playgroud)
这些使您免于模式匹配的麻烦,而无需依赖其他库。