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basic-types/statements-expressions #183
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basic-types/statements-expressions
Learning Rust By Practice, narrowing the gap between beginner and skilled-dev with challenging examples, exercises and projects.
https://zh.practice.rs/basic-types/statements-expressions.html
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👍 45 -
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👀 4
Replies: 80 comments 22 replies
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完成 3/3
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👍 4 -
❤️ 1
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nice practice!
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done
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🚀 1
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Done
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Done
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Done!在很多语言都有语句和表达式的概念!
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done
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assert_eq!((), {});为什么是可以通过断言的呢?
亲测,在第一题的assert_eq!里把"3"改成"{}"或者"()"都是可以的。
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assert_eq!() 的第二个参数是 {} ,是一个块作用域(也就是表达式),但是没有返回值,因此隐式地返回 ()
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👍 19 -
❤️ 1
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回答的好!
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Done
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x += 2
相当于
x = x + 2
虽然没有加 ; 但跑不起来,改成 x + 2 就行了,有点迷糊
没有加 ; 赋值操作也不算表达式么
fn main() {
let v = {
let mut x = 1;
x + 2
};
assert_eq!(v, 3);
}
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我的理解是 x += 2 和 x = x + 2 一样都是语句,而 x + 2 是表达式,语句块用不带分号的表达式结尾才会返回一个数值,用不带分号的语句和带分号的语句结尾在这里都一样,返回"()"
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以上理解似乎是错误的,x+=2好像也是表达式...
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👍 2
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是的,我的理解是,x+=2是表达式,没有返回值,所以{}返回默认()
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我也没懂,为啥x+=2就不行 x+2就行
想了想,感觉是x += 2已经返回赋值给了x然后v就不知道给什么了,x+2是出来了3然后就返回赋值给v
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总之,能返回值,它就是表达式
形似 x = 1 这种操作返回的是单元类型, 即是没有返回值的, 可以把它看做是表达式, 但结果就是返回单元类型 (), 为了区分语句和表达式, 也是为了优化性能(?存疑), 采用这种区分方式。
x += 2 和 x = x + 2 都是对右式进行求值, 在赋值左式, 最后的操作是赋值操作, 赋值操作是没有返回值的, 是一个语句, 如何把它看做表达式, 返回 ()
本人的简单理解, 如有不对, 请指正
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👍 1
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fn main() {
let v = {
let mut x = 1;
x += 2
};
assert_eq!(v, 3);
}
x=x+2这一句,按直觉理解应该是作为语句的,作为语句时应该加;则返回()。
但如果不加;,即是作为表达式,那么赋值语句的表达式可以返回(),我的理解对吗
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不是的,语句是没有返回值的,在the book里面有这样的一段:
Statements are instructions that perform some action and do not return a value.
我认为应该是如果加 ; 的话就是语句,是没有返回值的。而如果不加 ; 的话,就是一个表达式,它有返回值,但是赋值表达式的返回值是 ()
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👍 2
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fn main() {
let a = 8;
}
和
fn main() {
let a = 8
}
都是对的,说明rust 在只有一行语句的时候可以不写";" 但是下面有一行就会报错,所以不加";"的有可能是语句
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试了几种写法,觉得关键在mut上,因为mut是修改同一个内存地址上的值,并不会发生内存对象的再分配,所以最后x+=2才能作为表达式。但还是没有找到表达式和语句的本质区别相关的解释内容。
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done
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试了几种写法,觉得关键在mut上,因为mut是修改同一个内存地址上的值,并不会发生内存对象的再分配,所以最后x+=2才能作为表达式。但还是没有找到表达式和语句的本质区别相关的解释内容。
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不好意思,我是想回复上面一条,发现自己回复错了
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2022年11月9日 Done
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Done!
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Done
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Done
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mark
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细节太多..
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done
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谢谢大佬的教程
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mark
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done
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done!
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day1
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111
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Done!
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done
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done
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Done
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第三题提供另一种解法:
fn main() { let v = {let x = 3;}; assert!(v == ()); }
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✅
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Success(3/3)
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done
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其实3也可以有两种方法解决
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I'm the best
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