compound-types/tuple

[giscus[bot]]

compound-types/tuple

Learning Rust By Practice, narrowing the gap between beginner and skilled-dev with challenging examples, exercises and projects.

https://zh.practice.rs/compound-types/tuple.html

[roushanyyzz]

Done.

[paxtang21]

done!

[makexiao113]

Done.

[jan6055]

done

[colorsu]

Done.

[bruceblink]

第3题为什么12个元素可以打印，13个元素就不能打印，这是哪里规定的

[LiYpoi]

Why is tuple formatting limited to 12 items in Rust?
宏规定的https://doc.rust-lang.org/src/core/fmt/mod.rs.html#2008-2019

[ZLD1819858]

@LiYpoi，谢谢指引

[ZLD1819858]

Done.

[csx12588]

变量不可以只定义不绑定值，第五题，元组可以只定义，不绑定值吗？

[ytianxia6]

可以先定义，再绑定值，但不能不绑定值。

[guchen-yh]

Done!

[Grapeve]

2022.11.14 Done.

[cbldlpl]

done

[Kori000]

done

[denganjia]

done!

[XU0618]

done lala~

[Vocal-map]

done

[AllenWu233]

Done.

[TDAkory]

mark finished

[ty-zyp]

done

[SandyLi0404]

done

[lichenbuliren]

done

[chenmx00]

Done!

[XueHua-s]

Done, It's Easy

[Bogdanxin]

第五题：
答案给的是

fn main() {
    let (x, y, z);

    // 填空
    (y,z,x) = (1, 2, 3);
    
    assert_eq!(x, 3);
    assert_eq!(y, 1);
    assert_eq!(z, 2);
}

我自己没有按照要求填空也成功，但没明白其中的原因，是第一句的声明已经给元祖中的做变量绑定了吗？

fn main() {
    let (x, y, z): (i32, i32, i32);

    // 填空
    let (y,z,x) = (1, 2, 3);

    assert_eq!(x, 3);
    assert_eq!(y, 1);
    assert_eq!(z, 2);
}

[ytianxia6]

第一行 let (x, y, z): (i32, i32, i32);　　在你的代码里应该一点用都没有吧。你把它删除了应该一点影响都没有。

你这里两个 let ，第2个应该会重新定义了变量

[zhengkezhou1]

day3

[konlon-ya]

打卡

[FlandreLiao]

done

[bansal-harsh-2504]

✅ day 6

[knightArtorias777]

done!

[ryzen-xp]

Done >>>>>>>>>>>>>>>>>>

[ZaneLeo1]

Done.