如何实现自定义的“fmt::Debug"特征? [英] How to implement a custom 'fmt::Debug' trait?

问题描述

我猜你会这样做:

extern crate uuid;

use uuid::Uuid;
use std::fmt::Formatter;
use std::fmt::Debug;

#[derive(Debug)]
struct BlahLF {
    id: Uuid,
}

impl BlahLF {
    fn new() -> BlahLF {
        return BlahLF { id: Uuid::new_v4() };
    }
}

impl Debug for BlahLF {
    fn fmt(&self, &mut f: Formatter) -> Result {
        write!(f.buf, "Hi: {}", self.id);
    }
}

...但是尝试实现这个特性会产生:

...but attempting to implement this trait generates:

error[E0243]: wrong number of type arguments
  --> src/main.rs:19:41
   |
19 |     fn fmt(&self, &mut f: Formatter) -> Result {
   |                                         ^^^^^^ expected 2 type arguments, found 0

然而，这似乎是其他实现的方式.我做错了什么?

However, that seems to be how other implementations do it. What am I doing wrong?

推荐答案

根据 std::fmt 文档:

According to the example from the std::fmt docs:

extern crate uuid;

use uuid::Uuid;
use std::fmt;

struct BlahLF {
    id: Uuid,
}

impl fmt::Debug for BlahLF {
    fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
        write!(f, "Hi: {}", self.id)
    }
}

要强调的部分是fmt::Result中的fmt::.没有它，你指的是普通的 Result 类型.普通的 Result 类型确实有两个泛型类型参数，fmt::Result 没有.

The part to emphasize is the fmt:: in fmt::Result. Without that you're referring to the plain Result type. The plain Result type does have two generic type parameters, fmt::Result has none.

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