我想用to_string()
作为参数在Rust中创建一个&self
fn,并在函数内调用&self
元素的引用:
//! # Messages
//!
//! Module that builds and returns messages with user and time stamps.
use time::{Tm};
/// Represents a simple text message.
pub struct SimpleMessage<'a, 'b> {
pub moment: Tm,
pub content: &'b str,
}
impl<'a, 'b> SimpleMessage<'a, 'b> {
/// Gets the elements of a Message and transforms them into a String.
pub fn to_str(&self) -> String {
let mut message_string =
String::from("{}/{}/{}-{}:{} => {}",
&self.moment.tm_mday,
&self.moment.tm_mon,
&self.moment.tm_year,
&self.moment.tm_min,
&self.moment.tm_hour,
&self.content);
return message_string;
}
}
但是$ cargo run
回归:
error[E0061]: this function takes 1 parameter but 8 parameters were supplied
--> src/messages.rs:70:13
|
70 | / String::from("{}/{}/{}-{}:{}, {}: {}",
71 | | s.moment.tm_mday,
72 | | s.moment.tm_mon,
73 | | s.moment.tm_year,
... |
76 | | s.user.get_nick(),
77 | | s.content);
| |___________________________________^ expected 1 parameter
我真的不明白这种语法的问题,我错过了什么?
您可能打算使用format!
宏:
impl<'b> SimpleMessage<'b> {
/// Gets the elements of a Message and transforms them into a String.
pub fn to_str(&self) -> String {
let message_string =
format!("{}/{}/{}-{}:{} => {}",
&self.moment.tm_mday,
&self.moment.tm_mon,
&self.moment.tm_year,
&self.moment.tm_min,
&self.moment.tm_hour,
&self.content);
return message_string;
}
}
String::from
来自From
特征,它定义了一个from
方法,该方法采用单个参数(因此“此函数在错误消息中占用1个参数”)。
format!
已经生产了String
,因此无需转换。