167 lines
3.0 KiB
Markdown
167 lines
3.0 KiB
Markdown
# 数值类型
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### 整数
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1. 🌟
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> Tips: 如果我们没有显式的给予变量一个类型，那编译器会自动帮我们推导一个类型
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```rust,editable
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// 移除某个部分让代码工作
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fn main() {
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let x: i32 = 5;
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let mut y: u32 = 5;
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y = x;
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let z = 10; // 这里 z 的类型是?
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}
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```
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2. 🌟
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```rust,editable
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// 填空
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fn main() {
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let v: u16 = 38_u8 as __;
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}
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```
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3. 🌟🌟🌟
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> Tips: 如果我们没有显式的给予变量一个类型，那编译器会自动帮我们推导一个类型
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```rust,editable
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// 修改 `assert_eq!` 让代码工作
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fn main() {
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let x = 5;
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assert_eq!("u32".to_string(), type_of(&x));
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}
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// 以下函数可以获取传入参数的类型，并返回类型的字符串形式，例如 "i8", "u8", "i32", "u32"
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fn type_of<T>(_: &T) -> String {
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format!("{}", std::any::type_name::<T>())
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}
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```
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4. 🌟🌟
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```rust,editable
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// 填空，让代码工作
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fn main() {
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assert_eq!(i8::MAX, __);
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assert_eq!(u8::MAX, __);
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}
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```
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5. 🌟🌟
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```rust,editable
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// 解决代码中的错误和 `panic`
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fn main() {
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let v1 = 251_u8 + 8;
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let v2 = i8::checked_add(251, 8).unwrap();
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println!("{},{}",v1,v2);
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}
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```
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6. 🌟🌟
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```rust,editable
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// 修改 `assert!` 让代码工作
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fn main() {
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let v = 1_024 + 0xff + 0o77 + 0b1111_1111;
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assert!(v == 1579);
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}
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```
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### 浮点数
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7. 🌟
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```rust,editable
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// 将 ? 替换成你的答案
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fn main() {
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let x = 1_000.000_1; // ?
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let y: f32 = 0.12; // f32
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let z = 0.01_f64; // f64
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}
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```
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8. 🌟🌟 使用两种方法来让下面代码工作
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```rust,editable
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fn main() {
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assert!(0.1+0.2==0.3);
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}
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```
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### 序列Range
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9. 🌟🌟 两个目标: 1. 修改 `assert!` 让它工作 2. 让 `println!` 输出: 97 - 122
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```rust,editable
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fn main() {
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let mut sum = 0;
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for i in -3..2 {
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sum += i
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}
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assert!(sum == -3);
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for c in 'a'..='z' {
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println!("{}",c);
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}
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}
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```
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10. 🌟🌟
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```rust,editable
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// 填空
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use std::ops::{Range, RangeInclusive};
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fn main() {
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assert_eq!((1..__), Range{ start: 1, end: 5 });
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assert_eq!((1..__), RangeInclusive::new(1, 5));
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}
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```
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### 计算
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11. 🌟
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```rust,editable
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// 填空，并解决错误
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fn main() {
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// 整数加法
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assert!(1u32 + 2 == __);
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// 整数减法
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assert!(1i32 - 2 == __);
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assert!(1u8 - 2 == -1);
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assert!(3 * 50 == __);
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assert!(9.6 / 3.2 == 3.0); // error ! 修改它让代码工作
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assert!(24 % 5 == __);
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// 逻辑与或非操作
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assert!(true && false == __);
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assert!(true || false == __);
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assert!(!true == __);
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// 位操作
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println!("0011 AND 0101 is {:04b}", 0b0011u32 & 0b0101);
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println!("0011 OR 0101 is {:04b}", 0b0011u32 | 0b0101);
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println!("0011 XOR 0101 is {:04b}", 0b0011u32 ^ 0b0101);
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println!("1 << 5 is {}", 1u32 << 5);
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println!("0x80 >> 2 is 0x{:x}", 0x80u32 >> 2);
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}
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```
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> 你可以在[这里](https://github.com/sunface/rust-by-practice/blob/master/solutions/basic-types/numbers.md)找到答案(在 solutions 路径下)
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