对于实现相同特征的结构，如何克服具有不兼容类型的匹配臂?

问题描述

我正在尝试编写 cat 命令来学习 Rust，但我似乎无法将命令行参数转换为读取器结构.

I am attempting to write the cat command to learn Rust, but I can't seem to convert command line arguments into reader structs.

use std::{env, io};
use std::fs::File;

fn main() {
    for arg in env::args().skip(1) {
        let reader = match arg.as_str() {
            "-" => io::stdin(),
            path => File::open(&path).unwrap(),
        };
    }
}

错误:

error[E0308]: match arms have incompatible types
 --> src/main.rs:6:22
  |
6 |         let reader = match arg.as_str() {
  |                      ^ expected struct `std::io::Stdin`, found struct `std::fs::File`
  |
  = note: expected type `std::io::Stdin`
  = note:    found type `std::fs::File`
note: match arm with an incompatible type
 --> src/main.rs:8:21
  |
8 |             path => File::open(&path).unwrap(),
  |                     ^^^^^^^^^^^^^^^^^^^^^^^^^^

似乎不可能多态匹配特征实现者(相关).如何使用 File 或 Stdin 作为阅读器?

It doesn't seem like it's possible to polymorphically match trait implementers (related). How can I use either a File or Stdin as a reader?

推荐答案

问题在于 stdin() 返回一个类型为 Stdio 和 File 的对象::open(...).unwrap() 返回一个 File 类型的对象.在 Rust 中，匹配的所有分支都必须返回相同类型的值.

The problem is that stdin() returns an object of type Stdio and File::open(...).unwrap() returns an object of type File. In Rust, all arms of a match have to return values of the same type.

在这种情况下，您可能想要返回一个通用的 Read 对象.不幸的是 Read 是一个 trait，所以你不能按值传递它.最简单的替代方法是使用堆分配:

In this case you probably wanted to return a common Read object. Unfortunately Read is a trait so you cannot pass it by value. The easiest alternative is to resort to heap allocation:

use std::{env, io};
use std::io::prelude::*;
use std::fs::File;

fn main() {
    for arg in env::args().skip(1) {
        let reader = match arg.as_str() {
            "-" => Box::new(io::stdin()) as Box<Read>,
            path => Box::new(File::open(&path).unwrap()) as Box<Read>,
        };
    }
}

这篇关于对于实现相同特征的结构，如何克服具有不兼容类型的匹配臂?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！