rustling练习II–primitive type、move_semantic
rustling练习II–primitive type、move_semantic
用户7267083
发布于 2022-12-08 15:12:51
发布于 2022-12-08 15:12:51
rustling练习II–primitive type、move_semantic
于2022年10月14日2022年10月14日由Sukuna发布
4-1
// primitive_types1.rs
// Fill in the rest of the line that has code missing!
// No hints, there's no tricks, just get used to typing these :)
// I AM NOT DONE
fn main() {
// Booleans (`bool`)
let is_morning = true;
if is_morning {
println!("Good morning!");
}
let // Finish the rest of this line like the example! Or make it be false!
if is_evening {
println!("Good evening!");
}
}只用完成一行的代码就好了.不知道这题的意义是什么23333 就是写成let is_evening = false;就好了
4-2
// primitive_types2.rs
// Fill in the rest of the line that has code missing!
// No hints, there's no tricks, just get used to typing these :)
// I AM NOT DONE
fn main() {
// Characters (`char`)
// Note the _single_ quotes, these are different from the double quotes
// you've been seeing around.
let my_first_initial = 'C';
if my_first_initial.is_alphabetic() {
println!("Alphabetical!");
} else if my_first_initial.is_numeric() {
println!("Numerical!");
} else {
println!("Neither alphabetic nor numeric!");
}
let your_character = '1';// Finish this line like the example! What's your favorite character?
// Try a letter, try a number, try a special character, try a character
// from a different language than your own, try an emoji!
if your_character.is_alphabetic() {
println!("Alphabetical!");
} else if your_character.is_numeric() {
println!("Numerical!");
} else {
println!("Neither alphabetic nor numeric!");
}
}代码就是给定一个字符,进行判断而已.这题似乎也没难度,填写一个:let your_character = '1';就好
4-3
这题需要我们学会怎么声明一个数组,这个和C语言几乎一样.就是let a = [1,2,3,4];
4-4
这题需要我们创建一个数组slice,就是取第2-4个元素
// primitive_types4.rs
// Get a slice out of Array a where the ??? is so that the test passes.
// Execute `rustlings hint primitive_types4` or use the `hint` watch subcommand for a hint.
#[test]
fn slice_out_of_array() {
let a = [1, 2, 3, 4, 5];
let nice_slice = ???;
assert_eq!([2, 3, 4], nice_slice)
}答案是&a[1..4],这个slice:&a[a,b]会取a+1到b的元素.
4-5
这题需要我们写一个元组模式匹配的代码,就是(a,b)=元组
// primitive_types5.rs
// Destructure the `cat` tuple so that the println will work.
// Execute `rustlings hint primitive_types5` or use the `hint` watch subcommand for a hint.
// I AM NOT DONE
fn main() {
let cat = ("Furry McFurson", 3.5);
let /* your pattern here */ = cat;
println!("{} is {} years old.", name, age);
}把注释改成let (name,age) = cat;就好,这样可以像SML一样完成模式匹配
4-6
这题需要我们取出元组的第2个元素.
// primitive_types6.rs
// Use a tuple index to access the second element of `numbers`.
// You can put the expression for the second element where ??? is so that the test passes.
// Execute `rustlings hint primitive_types6` or use the `hint` watch subcommand for a hint.
// I AM NOT DONE
#[test]
fn indexing_tuple() {
let numbers = (1, 2, 3);
// Replace below ??? with the tuple indexing syntax.
let second = ???;
assert_eq!(2, second,
"This is not the 2nd number in the tuple!")
}答案就是let second = numbers.1;
5-1
// move_semantics1.rs
// Execute `rustlings hint move_semantics1` or use the `hint` watch subcommand for a hint.
fn main() {
let vec0 = Vec::new();
let vec1 = fill_vec(vec0);
println!("{} has length {} content `{:?}`", "vec1", vec1.len(), vec1);
vec1.push(88);
println!("{} has length {} content `{:?}`", "vec1", vec1.len(), vec1);
}
fn fill_vec(vec: Vec<i32>) -> Vec<i32> {
let mut vec = vec;
vec.push(22);
vec.push(44);
vec.push(66);
vec
}讲vec1改成mut(可变)即可,对Vec做增删改需要将Vec改为可变
5-2
// move_semantics2.rs
// Make me compile without changing line 13 or moving line 10!
// Execute `rustlings hint move_semantics2` or use the `hint` watch subcommand for a hint.
fn main() {
let vec0 = Vec::new();
let mut vec1 = fill_vec(vec0);
// Do not change the following line!
println!("{} has length {} content `{:?}`", "vec0", vec0.len(), vec0);
vec1.push(88);
println!("{} has length {} content `{:?}`", "vec1", vec1.len(), vec1);
}
fn fill_vec(vec: Vec<i32>) -> Vec<i32> {
let mut vec = vec;
vec.push(22);
vec.push(44);
vec.push(66);
vec
}vec0的所有权进入到函数fill_vec中了,fill_vec结束后,vec0的所有权被释放,main丧失了对vec0的所有权,就不能访问vec0了.
所以说我们要把传vec0改成传vec0的引用,传引用是不会交出所有权的.在函数中对此引用产生一个拷贝,将此拷贝传回来,这个拷贝是实际的Vec类型而不是引用
// move_semantics2.rs
// Make me compile without changing line 13 or moving line 10!
// Execute `rustlings hint move_semantics2` or use the `hint` watch subcommand for a hint.
fn main() {
let vec0 = Vec::new();
let mut vec1 = fill_vec(&vec0);
// Do not change the following line!
println!("{} has length {} content `{:?}`", "vec0", vec0.len(), vec0);
vec1.push(88);
println!("{} has length {} content `{:?}`", "vec1", vec1.len(), vec1);
}
fn fill_vec(vec: &Vec<i32>) -> Vec<i32> {
let mut vec = (*vec).to_vec();
vec.push(22);
vec.push(44);
vec.push(66);
(*vec).to_vec()
}5-3
fn main() {
let vec0 = Vec::new();
let mut vec1 = fill_vec(vec0);
println!("{} has length {} content `{:?}`", "vec1", vec1.len(), vec1);
vec1.push(88);
println!("{} has length {} content `{:?}`", "vec1", vec1.len(), vec1);
}
fn fill_vec(vec: Vec<i32>) -> Vec<i32> {
vec.push(22);
vec.push(44);
vec.push(66);
vec
}主要的问题就是fill_vec函数中的vec是不可变的,要push内容是不可以的.
一个解决方法就是创建一个拷贝,像5-1一样.
第二个方法就是将vec0设置成mut的,fill_vec的vec也设置成mut
fn main() {
let mut vec0 = Vec::new();
let mut vec1 = fill_vec(vec0);
println!("{} has length {} content `{:?}`", "vec1", vec1.len(), vec1);
vec1.push(88);
println!("{} has length {} content `{:?}`", "vec1", vec1.len(), vec1);
}
fn fill_vec(mut vec: Vec<i32>) -> Vec<i32> {
vec.push(22);
vec.push(44);
vec.push(66);
vec
}5-4
// move_semantics4.rs
// Refactor this code so that instead of passing `vec0` into the `fill_vec` function,
// the Vector gets created in the function itself and passed back to the main
// function.
// Execute `rustlings hint move_semantics4` or use the `hint` watch subcommand for a hint.
fn main() {
let vec0 = Vec::new();
let mut vec1 = fill_vec(vec0);
println!("{} has length {} content `{:?}`", "vec1", vec1.len(), vec1);
vec1.push(88);
println!("{} has length {} content `{:?}`", "vec1", vec1.len(), vec1);
}
// `fill_vec()` no longer takes `vec: Vec<i32>` as argument
fn fill_vec() -> Vec<i32> {
let mut vec = vec;
vec.push(22);
vec.push(44);
vec.push(66);
vec
}不用vec0就把vec0相关的内容删掉,改成在函数体内new一个就好
fn main() {
let mut vec1 = fill_vec();
println!("{} has length {} content `{:?}`", "vec1", vec1.len(), vec1);
vec1.push(88);
println!("{} has length {} content `{:?}`", "vec1", vec1.len(), vec1);
}
// `fill_vec()` no longer takes `vec: Vec<i32>` as argument
fn fill_vec() -> Vec<i32> {
let mut vec = Vec::new();
vec.push(22);
vec.push(44);
vec.push(66);
vec
}5-5
// move_semantics5.rs
// Make me compile only by reordering the lines in `main()`, but without
// adding, changing or removing any of them.
// Execute `rustlings hint move_semantics5` or use the `hint` watch subcommand for a hint.
// I AM NOT DONE
fn main() {
let mut x = 100;
let y = &mut x;
let z = &mut x;
*y += 100;
*z += 1000;
assert_eq!(x, 1200);
}不能对一个变量同时创建两个可变引用,改成这样就好:
fn main() {
let mut x = 100;
let z = &mut x;
*z += 100;
*z += 1000;
assert_eq!(x, 1200);
}5-6
// move_semantics6.rs
// Execute `rustlings hint move_semantics6` or use the `hint` watch subcommand for a hint.
// You can't change anything except adding or removing references.
fn main() {
let data = "Rust is great!".to_string();
get_char(data);
string_uppercase(&data);
}
// Should not take ownership
fn get_char(data: String) -> char {
data.chars().last().unwrap()
}
// Should take ownership
fn string_uppercase(mut data: &String) {
data = &data.to_uppercase();
println!("{}", data);
}代码的注释都写了,不要放弃所有权就是加引用,放弃所有权就是不加引用
fn main() {
let data = "Rust is great!".to_string();
get_char(&data);
string_uppercase(data);
}
// Should not take ownership
fn get_char(data: &String) -> char {
(*data).chars().last().unwrap()
}
// Should take ownership
fn string_uppercase(mut data: String) {
data = data.to_uppercase();
println!("{}", data);
}
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原始发表：2022年10月14日，如有侵权请联系 cloudcommunity@tencent.com 删除
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