为什么这个 Rust 程序忽略了不变性 [英] Why does this Rust program ignore immutability

问题描述

我有以下 Rust 程序，我希望它会导致编译错误，因为 x 稍后被重新分配.但它遵守并提供输出.为什么?

I have the following Rust program and I expect it to result in an compilation error since x is reassigned later. But it complies and gives output. Why?

fn main() {
   let (x, y) = (1, 3);
   println!("X is {} and Y is {}", x, y);

   let x: i32 = 565;
   println!("Now X is {}", x);
}

推荐答案

Rust 实际上可以让你隐藏块中的其他变量，所以 let x: i32 = 565; 定义了一个 new 变量 x 隐藏了之前定义的 let (x,y) = (1,3); 的 x.请注意，您甚至可以重新定义 x 以具有不同的类型，因为第二个 x 是一个全新的变量！

Rust actually lets you shadow other variables in a block, so let x: i32 = 565; is defining a new variable x that shadows the x defined earlier with let (x,y) = (1,3);. Note that you could even have redefined x to have a different type since the second x is a whole new variable!

fn main(){
   let x = 1;
   println!("Now X is {}",x);

   let x = "hi";
   println!("Now X is {}",x);
}

这个reddit 线程更详细地说明了为什么会这样有用.提到的两件事似乎很有趣:

This reddit thread goes into more detail about why this is useful. The two things that are mentioned that seem interesting are:

• 对于获取变量所有权但返回另一个相同类型的变量的操作，有时将返回的变量重新定义为具有相同的名称看起来不错".来自 此处:

let iter = vec.into_iter();
let iter = modify(iter);
let iter = double(iter);

或者使变量不可变:

Or to make a variable immutable:

let mut x;
// Code where `x` is mutable
let x = x;
// Code where `x` is immutable

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