ownership/borrowing #373
giscus[bot]
bot
Replies: 62 comments 15 replies
-
|
逐渐开始硬核起来了，烧脑 |
Beta Was this translation helpful? Give feedback.
-
|
done! lala~ |
Beta Was this translation helpful? Give feedback.
-
|
第6题，哪里tm讲了ref |
Beta Was this translation helpful? Give feedback.
-
|
推理猜一下（doge |
Beta Was this translation helpful? Give feedback.
-
|
5.1所有权习题里面提了一个部分Move。其中的示例用到了ref |
Beta Was this translation helpful? Give feedback.
-
|
你已经是一个成熟的程序员了，是时候自己猜了 |
Beta Was this translation helpful? Give feedback.
-
|
新编译器作用域在最后一次使用的位置 所以第10题 最后一句可以把 |
Beta Was this translation helpful? Give feedback.
-
|
我特意用1.70的编译器试了一遍，还真是 |
Beta Was this translation helpful? Give feedback.
-
|
这个第10题一下还真没想到 |
Beta Was this translation helpful? Give feedback.
-
|
打卡感謝課後練習 這樣才有學到的感覺 |
Beta Was this translation helpful? Give feedback.
-
|
不符合题目的一些内容 // 移除代码某个部分，让它工作
// 你不能移除整行的代码！
fn main() {
let mut s = String::from("hello");
let r1 = &mut s;
let r2 = &r1;
println!("{}, {}", r1, r2);
}10 // 注释掉一行代码让它工作
fn main() {
let mut s = String::from("hello, ");
let r1 = &mut s;
r1.push_str("world");
//r2.push_str("!");
println!("{}",r1);
let r2 = &mut s;
} |
Beta Was this translation helpful? Give feedback.
-
|
两个题目都是没问题的，你可以仔细看下： 7-不可变引用可以存在多个 // 移除代码某个部分，让它工作
// 你不能移除整行的代码！
fn main() {
let mut s = String::from("hello");
let r1 = &s;
let r2 = &s;
println!("{}, {}", r1, r2);
}10- 移除最后一行打印的代码，会让r1的作用域仅仅到r1.push就结束了，不违反某一作用域内同时存在多个可变引用。 |
Beta Was this translation helpful? Give feedback.
-
|
1、第一题，引用的值 Q: fn main() {
let x = 5;
// 填写空白处
let p = __;
println!("x 的内存地址是 {:p}", p); // output: 0x16fa3c84
}A: fn main() {
let x = 5;
let p = &x;//获取指向x的地址
println!("x 的内存地址是 {:p}", p); // output: 0x16fa3ac84
}2、第二题，如何解引用 Q: fn main() {
let x = 5;
let y = &x;
// 只能修改以下行
assert_eq!(5, y);
}A: fn main() {
let x = 5;
let y = &x;
// 只能修改以下行
assert_eq!(5, *y);//解引用获取引用y指向地址的值，也就是x
}3、第三题，修复错误 Q: // 修复错误
fn main() {
let mut s = String::from("hello, ");
borrow_object(s)
}
fn borrow_object(s: &String) {}A: fn main() {
let mut s = String::from("hello, ");
borrow_object(&s)//只需要求改这里即可
}
fn borrow_object(s: &String) {}//其实题干已经提示了，在这里要传入的是一个引用，引用不触发所有权矛盾4、第四题，可变引用 Q: // 修复错误
fn main() {
let mut s = String::from("hello, ");
push_str(s)
}
fn push_str(s: &mut String) {
s.push_str("world")
}A: fn main() {
let mut s = String::from("hello, ");
push_str(& mut s)//可以从函数调用中传参的类型来推理此处变量类型
}
fn push_str(s: &mut String) {
s.push_str("world")
}5、第五题 Q: fn main() {
let mut s = String::from("hello, ");
// 填写空白处，让代码工作
let p = __;
p.push_str("world");
}A: fn main() {
let mut s = String::from("hello, ");
// 填写空白处，让代码工作
let p = &mut s;//p传入的是指向s的地址，且因为下面有要求修改值，所以借用不够，还需要声明可变mut
p.push_str("world");
}结合上一节内容，考虑这里写成深度拷贝，但是如此以来修改部分就不止 fn main() {
let mut s = String::from("hello, ");
let mut p = s.clone();
p.push_str("world");
}6、第六题，ref和& Q: fn main() {
let c = '中';
let r1 = &c;
// 填写空白处，但是不要修改其它行的代码
let __ r2 = c;
assert_eq!(*r1, *r2);
// 判断两个内存地址的字符串是否相等
assert_eq!(get_addr(r1),get_addr(r2));
}
// 获取传入引用的内存地址的字符串形式
fn get_addr(r: &char) -> String {
format!("{:p}", r)
}A: fn main() {
let c = '中';
let r1 = &c;
// 我浅薄理解为：ref在左边使用，&在右边使用
let ref r2 = c;
assert_eq!(*r1, *r2);
// 判断两个内存地址的字符串是否相等
assert_eq!(get_addr(r1),get_addr(r2));
}
// 获取传入引用的内存地址的字符串形式
fn get_addr(r: &char) -> String {
format!("{:p}", r)
}7、第七题，可变引用以及多个引用 Q: // 移除代码某个部分，让它工作
// 你不能移除整行的代码！
fn main() {
let mut s = String::from("hello");
let r1 = &mut s;
let r2 = &mut s;
println!("{}, {}", r1, r2);
}A: fn main() {
let mut s = String::from("hello");
//一个域里面有一个可变引用后就无法使用其他引用，所以mut必须都去掉
let r1 = & s;
let r2 = & s;
println!("{}, {}", r1, r2);
}8、第八题，从不可变对象中借用可变 Q: fn main() {
// 通过修改下面一行代码来修复错误
let s = String::from("hello, ");
borrow_object(&mut s)
}
fn borrow_object(s: &mut String) {}A: fn main() {
// String属于不可变对象？又学到了，然后mut自然就不可以用了
let s = String::from("hello, ");
borrow_object(& s)
}
fn borrow_object(s: & String) {}10、第十题 Q: // 注释掉一行代码让它工作
fn main() {
let mut s = String::from("hello, ");
let r1 = &mut s;
r1.push_str("world");
let r2 = &mut s;
r2.push_str("!");
println!("{}",r1);
}A: // 注释掉一行代码让它工作
fn main() {
let mut s = String::from("hello, ");
let r1 = &mut s;
r1.push_str("world");//调用完这个函数之后，r1就被释放掉了
let r2 = &mut s;
r2.push_str("!");
//println!("{}",r1);
}11、第十一题 Q: fn main() {
let mut s = String::from("hello, ");
let r1 = &mut s;
let r2 = &mut s;
// 在下面增加一行代码人为制造编译错误：cannot borrow `s` as mutable more than once at a time
// 你不能同时使用 r1 和 r2
}A: fn main() {
let mut s = String::from("hello, ");
let r1 = &mut s;
let r2 = &mut s;
r1.push_str("world");//因为一次只有一个可变引用可以改值，所以添加这个编译器就会报错（好家伙，我以为不添加这一行的话按照我的理解也会报错，原来还是需要尝试通过引用改变来触发这个编译error）
} |
Beta Was this translation helpful? Give feedback.
-
|
第十题 r1在调用函数后并没有释放，不然你在下一句重新再用r1调用试试。因为又绑定了r2为s的可变引用，所以r1才被释放的。 |
Beta Was this translation helpful? Give feedback.
-
|
不是，是因为nll发现r1在r2声明之前就不再使用了，所以编译器直接drop了r1 |
Beta Was this translation helpful? Give feedback.
-
|
是你，破墙侠！！！！，我老是访问不到gitup!!! |
Beta Was this translation helpful? Give feedback.
-
|
第5题如果不按划线的话，也可以写成 或 |
Beta Was this translation helpful? Give feedback.
-
|
前面的教程中好像没有对ref的介绍 |
Beta Was this translation helpful? Give feedback.
-
|
mark finished |
Beta Was this translation helpful? Give feedback.
-
|
第9题不用改任何东西~ |
Beta Was this translation helpful? Give feedback.
-
|
为啥输出这两个值一样的 fn main() {
let x = 5;
let p = &x;
println!("{}",*p );
println!("{}", p );
} |
Beta Was this translation helpful? Give feedback.
-
|
day2 完成所有权部分 |
Beta Was this translation helpful? Give feedback.
-
|
第10题其实是注释掉println!("{}",r1); // 注释掉一行代码让它工作
fn main() {
let mut s = String::from("hello, ");
let r1 = &mut s;
r1.push_str("world");
let r2 = &mut s;
r2.push_str("!");
// println!("{}",r1);
} |
Beta Was this translation helpful? Give feedback.
-
|
因为 r1 是一个可变引用,当我们创建 r2 时, r1 的生命周期应该已经结束了 |
Beta Was this translation helpful? Give feedback.
and others
-
ownership/borrowing
Learning Rust By Practice, narrowing the gap between beginner and skilled-dev with challenging examples, exercises and projects.
https://zh.practice.rs/ownership/borrowing.html
Beta Was this translation helpful? Give feedback.
All reactions