我正在按照这个参考来实现一个计算最小输入数的简单程序：
use std::io::prelude::*;
use std::io;
fn read_vec() -> Vec<i32> {
let mut vec: Vec<i32> = Vec::<i32>::new();
let stdin = io::stdin();
println!("Enter a list of numbers, one per line. End with Ctrl-D (Linux) or Ctrl-Z (Windows).");
for line in stdin.lock().lines() {
let line = line.unwrap();
match line.trim().parse::<i32>() {
Ok(num) => vec.push(num),
Err(_) => println!("What did I say about numbers?"),
}
}
vec
}
pub enum SomethingOrNothing<T> {
Something(T),
Nothing,
}
pub use self::SomethingOrNothing::*;
type NumberOrNothing = SomethingOrNothing<i32>;
pub trait Minimum: Copy {
fn min(self, b: Self) -> Self;
}
pub fn vec_min<T: Minimum>(v: Vec<T>) -> SomethingOrNothing<T> {
let mut min = Nothing;
for e in v {
min = match min {
Something(t) => Something(e.min(t)),
Nothing => Something(e),
}
}
min
}
impl Minimum for i32 {
fn min (self, b: Self) -> Self {
if self < b {self} else {b}
}
}
impl NumberOrNothing {
pub fn print(self) {
match self {
Nothing => println!("The number is: <nothing>"),
Something(n) => println!("{}", n),
};
}
}
fn main() {
let vec = read_vec();
let min = vec_min(vec);
min.print();
}
构建运行程序：
Enter a list of numbers, one per line. End with Ctrl-D (Linux) or Ctrl-Z (Windows).
100
8
200
8D
我们可以看到D在最小数字之后有一个尾随的“”：8。但是，如果我将输出更改为：
Something(n) => println!("{}", n),
至：
Something(n) => println!("The number is: {}", n),
输出看起来很正常：
Enter a list of numbers, one per line. End with Ctrl-D (Linux) or Ctrl-Z (Windows).
100
8
200
The number is: 8
我猜这个问题与标准输出缓冲区有关，但不知道为什么。任何人都可以提供一些线索吗？
PS，我可以在macOS( zsh) 和OmniOS( bash) 上重现此问题，但无法在Linux( bash) 上重现。