[英]Why is the trait bound Trait<&String> not satisfied even though I've implemented Trait<&str>?
I have written the following code: 我写了以下代码:
pub struct Serializer;
pub trait Serialize<T> {
fn to_bert(&self, data: T) -> Vec<u8>;
}
pub trait Convert<T> {
fn to_binary(&self, data: T) -> Vec<u8>;
}
impl<'a> Convert<&'a str> for Serializer {
fn to_binary(&self, data: &'a str) -> Vec<u8> {
let binary_string = data.as_bytes();
let binary_length = binary_string.len() as i16;
let mut binary = vec![];
binary.write_i16::<BigEndian>(binary_length).unwrap();
binary.extend(binary_string.iter().clone());
binary
}
}
impl Serialize<String> for Serializer {
fn to_bert(&self, data: String) -> Vec<u8> {
let binary_string = self.to_binary(&data);
self.generate_term(BertTag::String, binary_string)
}
}
impl<'a> Serialize<&'a str> for Serializer {
fn to_bert(&self, data: &'a str) -> Vec<u8> {
let binary_string = self.to_binary(data);
self.generate_term(BertTag::String, binary_string)
}
}
When compiling, I receive an error which says the compiler cannot find the right function for a call: 编译时,我收到一条错误消息,提示编译器找不到适合调用的函数:
error: the trait bound `serializers::Serializer: serializers::Convert<&std::string::String>` is not satisfied [E0277]
let binary_string = self.to_binary(&data);
^~~~~~~~~
help: run `rustc --explain E0277` to see a detailed explanation
help: the following implementations were found:
help: <serializers::Serializer as serializers::Convert<&'a str>>
help: <serializers::Serializer as serializers::Convert<types::BertType>>
Why can the compiler not find the right implementation when I've specified &str
with a lifetime? 当我指定了生命周期的
&str
时,为什么编译器找不到正确的实现? How can I fix this? 我怎样才能解决这个问题?
How can I fix this?
我怎样才能解决这个问题?
By getting a &str
from data
: 通过从
data
获取&str
:
let binary_string = self.to_binary(&*data);
Like the error says, Convert<&String>
is not implemented for Serializer
, but Convert<&str>
is. 就像错误所说的那样,
Convert<&String>
不是为Serializer
实现的,而Convert<&str>
是。 &data
is a &String
, but you need a &str
, writing &*data
produces a &str
. &data
是一个&String
,但是您需要一个&str
,编写&*data
会产生一个&str
。
Note that in general (when the string is not changed) it should not be necessary to implement the same functionality for both String
and &str
, only the &str
version should be enough. 请注意,通常(当不更改字符串时)不必为
String
和&str
实现相同的功能,只需&str
版本就足够了。 In your example, you can remove the Serialize<String>
implementation for Serializer
and use Serialize<&str>
instead. 在你的榜样,您可以删除
Serialize<String>
实施Serializer
并使用Serialize<&str>
代替。
fn main() {
let mut s = Serializer;
let a: String = "a".to_string();
// autoderef &String to &str and call Serialize<&str>::to_bert()
// I don't know why the autoderef does not work in your original example
s.to_bert(&a);
}
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