我正在尝试通过Rust by Example website上的“元组课程”,但我仍然停留在格式化的输出实现上。我有这个代码,它打印传递的矩阵:
<input-block ng-model="name" label="firstname" modern-style="true" name="firstname" type="text">
</input-block>
但编译器打印下一条消息:
#[derive(Debug)]
struct Matrix{
data: Vec<Vec<f64>> // [[...], [...],]
}
impl fmt::Display for Matrix {
fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
let output_data = self.data
// [[1, 2], [2, 3]] -> ["1, 2", "2, 3"]
.into_iter()
.map(|row| {
row.into_iter()
.map(|value| value.to_string())
.collect::<Vec<String>>()
.join(", ")
})
.collect::<Vec<String>>()
// ["1, 2", "2, 3"] -> ["(1, 2)", "(2, 3)"]
.into_iter()
.map(|string_row| { format!("({})", string_row) })
// ["(1, 2)", "(2, 3)"] -> "(1, 2),\n(2, 3)"
.collect::<Vec<String>>()
.join(",\n");
write!(f, "{}", output_data)
}
}
我试图将<anon>:21:40: 21:44 error: cannot move out of borrowed content [E0507]
<anon>:21 let output_data = self.data
^~~~
<anon>:21:40: 21:44 help: see the detailed explanation for E0507
error: aborting due to previous error
playpen: application terminated with error code 101
的结果包装到output_data
中,但编译器仍然会打印此错误。如何解决此问题,以便RefCell
宏正常工作?
答案 0 :(得分:5)
问题是,into_inter
取得data
的所有权,即从data
移出self
,这是不允许的(这就是错误说)。要在不获取所有权的情况下迭代向量,请使用iter
方法:
fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
let output_data = self.data
// [[1, 2], [2, 3]] -> ["1, 2", "2, 3"]
.iter()
.map(|row| {
row.into_iter()
.map(|value| value.to_string())
.collect::<Vec<String>>()
.join(", ")
})
.collect::<Vec<String>>()
// ["1, 2", "2, 3"] -> ["(1, 2)", "(2, 3)"]
.into_iter()
.map(|string_row| { format!("({})", string_row) })
// ["(1, 2)", "(2, 3)"] -> "(1, 2),\n(2, 3)"
.collect::<Vec<String>>()
.join(",\n");
write!(f, "{}", output_data)
}
看看Formatter。它有一些方法可以帮助编写fmt
。这是一个不分配中介向量和字符串的版本:
fn fmt(&self, f: &mut fmt::Formatter) -> fmt::Result {
let mut sep = "";
for line in &self.data { // this is like: for line in self.data.iter()
try!(f.write_str(sep));
let mut d = f.debug_tuple("");
for row in line {
d.field(row);
}
try!(d.finish());
sep = ",\n";
}
Ok(())
}