如何在不知道特定 IP 版本的情况下创建“IpAddr"? [英] How to create an `IpAddr` without knowing the specific IP version?

问题描述

我正在玩 Iron，但遇到了这个问题.

I am playing with Iron, and I ran into this problem.

fn main() {
    let mut router = Router::new();
    let address = "127.0.0.1"; // or maybe "::1/128"
    let port = 3000;
    let ip = std::net::IpAddr::new(address); // does not exist

    Iron::new(router).http((ip, port)).unwrap();
}

http() 方法采用一个结构体来实现 ToSocketAddrs.(&str, u16) 实现了这个特征，但我更喜欢在 http() 方法被调用之前验证用户输入的有效性.

The http() method takes a struct that implements ToSocketAddrs. (&str, u16) implements this trait, but I prefer to verify the validity of user input before the http() method is called.

我看到 (std::net::IpAddr, u16) 实现了这个特性，但我不知道如何不可知"地构建一个 IpAddr:也许用户写了一个 IPv4 地址，也许是 IPv6.

I saw that (std::net::IpAddr, u16) implements this trait, but I do not know how to build an IpAddr "agnostically": maybe the user wrote an IPv4 address, maybe an IPv6.

有没有办法仅从字符串创建 IpAddr ?我认为这是可能的，因为我可以给它一个 (&str, u16).

Is there a way to create an IpAddr from a string only? I think that it is possible because I can give to it a (&str, u16).

推荐答案

你的朋友是 FromStr 来自标准库的特征.它抽象了可以从字符串创建的类型.正如你所看到的 Ipv4Addr，Ipv6Addr 和 IpAddr 都实现了这个特性！所以你可以写:

Your friend is the FromStr trait from the standard library. It abstracts types that can be created from a string. As you can see Ipv4Addr, Ipv6Addr and IpAddr all implement that trait! So you could either write:

use std::str::FromStr;

let addr = IpAddr::from_str("127.0.0.1");

或者，稍微更常见的方法，通过使用 <代码>str::parse() 方法:

Or, the slightly more common way, by using the str::parse() method:

let addr = "127.0.0.1".parse::<IpAddr>();

from_str()/parse() 方法返回一个 Result 来表示字符串是否有效.

The from_str()/parse() methods return a Result to signal whether or not the string is valid.

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