collections/vector #214
giscus[bot]
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Replies: 22 comments 4 replies
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v2.extend(v1.iter()); |
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v2.extend(&v1); |
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好多都没学过 ，我不知道怎么做 |
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+1 |
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+1 |
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在练中学,在学中练 |
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2022.12.6 发现 solution 和题目对不上，这边补充下第三题的答案（只能保证可以跑，仅供参考） // 填空
fn main() {
// array -> Vec
// impl From<[T; N]> for Vec
let arr = [1, 2, 3];
let v1 = Vec::from(arr);
let v2: Vec<i32> = arr.into();
assert_eq!(v1, v2);
// String -> Vec
// impl From<String> for Vec
let s = "hello".to_string();
let v1: Vec<u8> = s.into();
let s = "hello".to_string();
let v2 = s.into_bytes();
assert_eq!(v1, v2);
// impl<'_> From<&'_ str> for Vec
let s = "hello";
let v3 = Vec::from(s);
assert_eq!(v2, v3);
// 迭代器 Iterators 可以通过 collect 变成 Vec
let v4: Vec<i32> = [0; 10].iter().map(|x| *x).collect();
assert_eq!(v4, vec![0; 10]);
println!("Success!")
} |
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第五题 // 修复错误
fn main() {
let mut v = vec![1, 2, 3];
let slice1 = &v[..];
// 越界访问将导致 panic.
// 修改时必须使用 `v.len`
let slice2 = &v[0..v.len()];
assert_eq!(slice1, slice2);
// 切片是只读的
// 注意：切片和 `&Vec` 是不同的类型，后者仅仅是 `Vec` 的引用，并可以通过解引用直接获取 `Vec`
let vec_ref: &mut Vec<i32> = &mut v;
(*vec_ref).push(4);
let mut slice3: Vec<i32> = Vec::from(&v[..3]);
slice3.push(4);
assert_eq!(slice3, &[1, 2, 3, 4]);
println!("Success!")
}
|
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第四题我的做法。忘了用if let了，会比match更简洁。 fn main() {
let mut v = Vec::from([1, 2, 3]);
for i in 0..5 {
match v.get(i)
{
Some(num)=>{
println!("{:?}", num);
v[i]=num+1;
},
None=>v.push(i+2)
}
}
assert_eq!(v, vec![2, 3, 4, 5, 6]);
println!("Success!")
}` |
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1、回顾一下对于 fn main() {
let arr: [u8; 3] = [1, 2, 3];
let v = Vec::from(arr);
is_vec(v.clone());
let v = vec![1, 2, 3];
is_vec(v.clone());
// vec!(..) 和 vec![..] 是同样的宏，宏可以使用 []、()、{}三种形式，因此...
let v = vec!(1, 2, 3);
is_vec(v.clone());
// ...在下面的代码中, v 是 Vec<[u8; 3]> , 而不是 Vec<u8>
// 使用 Vec::new 和 `for` 来重写下面这段代码
let mut v1=Vec::new();
for a in arr{
v1.push(a);
}
is_vec(v1.clone());
assert_eq!(v, v1);
println!("Success!")
}
fn is_vec(v: Vec<u8>) {}2、 fn main() {
let mut v1 = Vec::from([1, 2, 4]);
v1.pop();
v1.push(3);
let mut v2 = Vec::new();
v2.extend(v1.clone());//这里不clone的话会导致v1失去所有权
assert_eq!(v1, v2);
println!("Success!")
}3、 fn main() {
// array -> Vec
// impl From<[T; N]> for Vec
let arr = [1, 2, 3];
let v1 = Vec::from(arr);
let v2: Vec<i32> = arr.into();
assert_eq!(v1, v2);
// String -> Vec
// impl From<String> for Vec
let s = "hello".to_string();
let v1: Vec<u8> = s.into();
let s = "hello".to_string();
let v2 = s.into_bytes();
assert_eq!(v1, v2);
// impl<'_> From<&'_ str> for Vec
let s = "hello";
let v3 = Vec::from(s);
assert_eq!(v2, v3);
// 迭代器 Iterators 可以通过 collect 变成 Vec
let v4: Vec<i32> = [0; 10].into_iter().collect();
assert_eq!(v4, vec![0; 10]);
println!("Success!")
}4、索引 fn main() {
let mut v = Vec::from([1, 2, 3]);
for i in 0..3 {
println!("{:?}", v[i])
}
for i in 0..5 {
// 实现这里的代码...
if i<=2{
v[i]+=1;
}else{
v.push(i+2);
}
}
assert_eq!(v, vec![2, 3, 4, 5, 6]);
println!("Success!")
}5、 fn main() {
let mut v = vec![1, 2, 3];
let slice1 = &v[..];
// 越界访问将导致 panic.
// 修改时必须使用 `v.len`
let slice2 = &v[0..v.len()];
assert_eq!(slice1, slice2);
// 切片是只读的
// 注意：切片和 `&Vec` 是不同的类型，后者仅仅是 `Vec` 的引用，并可以通过解引用直接获取 `Vec`
let vec_ref: &mut Vec<i32> = &mut v;
(*vec_ref).push(4);
let slice3 = &vec_ref[0..4];
//slice3.push(4);
assert_eq!(slice3, &[1, 2, 3, 4]);
println!("Success!")
}6、 fn main() {
let mut vec = Vec::with_capacity(10);
assert_eq!(vec.len(), 0);
assert_eq!(vec.capacity(), 10);
// 由于提前设置了足够的容量，这里的循环不会造成任何内存分配...
for i in 0..10 {
vec.push(i);
}
assert_eq!(vec.len(), 10);
assert_eq!(vec.capacity(), 10);
// ...但是下面的代码会造成新的内存分配
vec.push(11);
assert_eq!(vec.len(), 11);
assert!(vec.capacity() >= 11);
// 填写一个合适的值，在 `for` 循环运行的过程中，不会造成任何内存分配
let mut vec = Vec::with_capacity(100);
for i in 0..100 {
vec.push(i);
}
assert_eq!(vec.len(),100);
assert_eq!(vec.capacity(), 100);
println!("Success!")
}7、用枚举对象在Vec中存储不同的类型 #[derive(Debug,PartialEq)]//注意此处需要实现PartialEq
enum IpAddr {
V4(String),
V6(String),
}
fn main() {
// 填空
let v : Vec<IpAddr>= Vec::from([
IpAddr::V4("127.0.0.1".to_string()),
IpAddr::V6("::1".to_string())
]);
// 枚举的比较需要派生 PartialEq 特征
assert_eq!(v[0], IpAddr::V4("127.0.0.1".to_string()));
assert_eq!(v[1], IpAddr::V6("::1".to_string()));
println!("Success!")
}8、实现同一个特征的可以用 trait IpAddr {
fn display(&self);
}
struct V4(String);
impl IpAddr for V4 {
fn display(&self) {
println!("ipv4: {:?}",self.0)
}
}
struct V6(String);
impl IpAddr for V6 {
fn display(&self) {
println!("ipv6: {:?}",self.0)
}
}
fn main() {
// 填空
let v: Vec<Box<dyn IpAddr>>= vec![
Box::new(V4("127.0.0.1".to_string())),
Box::new(V6("::1".to_string())),
];
for ip in v {
ip.display();
}
} |
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到这期确实变难了不少。 |
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This will produce error: Solutions:
Or:
|
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mark 11.2 5 |
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mark finished |
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Day 9 |
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第一题 fn main() {
let arr: [u8; 3] = [1, 2, 3];
let v = Vec::from(arr);
is_vec(&v);
let v = vec![1, 2, 3];
is_vec(&v);
// vec!(..) 和 vec![..] 是同样的宏，宏可以使用 []、()、{}三种形式，因此...
let v = vec!(1, 2, 3);
is_vec(&v);
// ...在下面的代码中, v 是 Vec<[u8; 3]> , 而不是 Vec<u8>
// 使用 Vec::new 和 `for` 来重写下面这段代码
// let v1 = vec!(arr);
let mut v1 = Vec::new();
for i in &arr {
v1.push(*i);
}
is_vec(&v1);
assert_eq!(v, v1);
println!("Success!")
}
fn is_vec(v: &Vec<u8>) {}第二题 // 填空
fn main() {
let mut v1 = Vec::from([1, 2, 4]);
v1.pop();
v1.push(3);
let mut v2 = Vec::new();
v2.extend([1,2,3]);
assert_eq!(v1, v2);
println!("Success!")
}第三题 // 填空
fn main() {
// array -> Vec
// impl From<[T; N]> for Vec
let arr = [1, 2, 3];
let v1 = Vec::from(arr);
let v2: Vec<i32> = arr.to_vec();
assert_eq!(v1, v2);
// String -> Vec
// impl From<String> for Vec
let s = "hello".to_string();
let v1: Vec<u8> = s.into();
let s = "hello".to_string();
let v2 = s.into_bytes();
assert_eq!(v1, v2);
// impl<'_> From<&'_ str> for Vec
let s = "hello";
let v3 = Vec::from(s);
assert_eq!(v2, v3);
// 迭代器 Iterators 可以通过 collect 变成 Vec
let v4: Vec<i32> = [0; 10].into_iter().collect();
assert_eq!(v4, vec![0; 10]);
println!("Success!")
}第四题 // 修复错误并实现缺失的代码
fn main() {
let mut v = Vec::from([1, 2, 3]);
for i in 0..3 {
println!("{:?}", v[i])
}
v.clear();
for i in 2..7 {
// 实现这里的代码...
v.push(i);
}
assert_eq!(v, vec![2, 3, 4, 5, 6]);
println!("Success!")
}第五题 // 修复错误
fn main() {
let mut v = vec![1, 2, 3];
let slice1 = &v[..];
// 越界访问将导致 panic.
// 修改时必须使用 `v.len`
let slice2 = &v[0..v.len()];
assert_eq!(slice1, slice2);
// 切片是只读的
// 注意：切片和 `&Vec` 是不同的类型，后者仅仅是 `Vec` 的引用，并可以通过解引用直接获取 `Vec`
let vec_ref: &mut Vec<i32> = &mut v;
(*vec_ref).push(4);
let slice3 = &mut Vec::from(&v[0..3]);
slice3.push(4);
assert_eq!(slice3, &[1, 2, 3, 4]);
println!("Success!")
}第六题 // 修复错误
fn main() {
let mut vec = Vec::with_capacity(10);
assert_eq!(vec.len(), 0);
assert_eq!(vec.capacity(), 10);
// 由于提前设置了足够的容量，这里的循环不会造成任何内存分配...
for i in 0..10 {
vec.push(i);
}
assert_eq!(vec.len(), 10);
assert_eq!(vec.capacity(), 10);
// ...但是下面的代码会造成新的内存分配
vec.push(11);
assert_eq!(vec.len(), 11);
assert!(vec.capacity() >= 11);
// 填写一个合适的值，在 `for` 循环运行的过程中，不会造成任何内存分配
let mut vec = Vec::with_capacity(101);
for i in 0..100 {
vec.push(i);
}
assert_eq!(vec.len(), 100);
assert_eq!(vec.capacity(), 101);
println!("Success!")
}第七题 #[derive(Debug, PartialEq)]
enum IpAddr {
V4(String),
V6(String),
}
fn main() {
// 填空
let v : Vec<IpAddr>= vec![
IpAddr::V4("127.0.0.1".to_string()),
IpAddr::V6("::1".to_string()),
];
// 枚举的比较需要派生 PartialEq 特征
assert_eq!(v[0], IpAddr::V4("127.0.0.1".to_string()));
assert_eq!(v[1], IpAddr::V6("::1".to_string()));
println!("Success!")
}第八题 trait IpAddr {
fn display(&self);
}
struct V4(String);
impl IpAddr for V4 {
fn display(&self) {
println!("ipv4: {:?}",self.0)
}
}
struct V6(String);
impl IpAddr for V6 {
fn display(&self) {
println!("ipv6: {:?}",self.0)
}
}
fn main() {
// 填空
let v: Vec<Box<dyn IpAddr>>= vec![
Box::new(V4("127.0.0.1".to_string())),
Box::new(V6("::1".to_string())),
];
for ip in v {
ip.display();
}
} |
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collections/vector
Learning Rust By Practice, narrowing the gap between beginner and skilled-dev with challenging examples, exercises and projects.
https://zh.practice.rs/collections/vector.html
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