我可以使向量可变
pub struct Test<'a>{
vec: &'a mut Vec<i32>,
}
impl<'a> Test<'a> {
pub fn created()->Test<'a>{
Test {vec: &'a mut Vec::new() }
}
pub fn add(&self, value: i32){
self.vec.push(value);
}
}
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expected `:`, found `mut`
Test {vec: &'a mut Vec::new() }
^~~
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这是一个类似的问题,但
答案就可以了,但是如果我不想要怎么办,可以执行“应用响应链接”
pub struct Test{
vec: Vec<i32>,
}
impl Test {
pub fn created()->Test {
Test {vec: Vec::new() }
}
pub fn add(&mut self, value: i32){
self.vec.push(value);
}
}
..//
let mut test: my::Test = my::Test::created();
test.add(1i32);
let mut test1: my::Test = my::Test::created();
test1 = test; <-- I do not want, you can do this
..//
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因为我可以使向量可变,而不必使其成为所有结构
也许您正在寻找内部可变性。请不要随意使用内部可变性,请先阅读此内容。
use std::cell::RefCell;
pub struct Test{
vec: RefCell<Vec<i32>>,
}
impl Test {
pub fn created()->Test {
Test {vec: RefCell::new(Vec::new()) }
}
pub fn add(&self, value: i32){
self.vec.borrow_mut().push(value);
}
}
fn main() {
let test = Test::created();
test.add(1i32);
let test1 = Test::created();
// test1 = test; // does not work anymore
}
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