Rust 中星号符号的用法是什么? [英] What is the usage of the asterisk symbol in Rust?

问题描述

我是 Rust 的新手，我不理解以下代码:

I'm new to Rust and I don't understand the following piece of code:

let mut x = 5;
{
    let y = &mut x;
    *y += 1;
}
println!("{}", x);

来自 Rust 站点的解释:

您还会注意到我们在 y 前面添加了一个星号 (*)，使其成为 *y，这是因为 y 是 &mut 引用.您还需要使用星号 [sic] 来访问参考文献的内容.

You'll also notice we added an asterisk (*) in front of y, making it *y, this is because y is a &mut reference. You'll need to use astrisks [sic] to access the contents of a reference as well.

如果*y是一个引用，为什么下面的代码有效

If *y is a reference, why does the following code work

fn main() {
    let mut x = 5;
    {
        let y = &mut x;
        println!("{}", y);
    }
}

我知道我没有修改这里的值，但有什么区别以及为什么y += 1; 会不起作用吗?

I know I'm not modifying the value here, but what is the difference and why would y += 1; not work?

推荐答案

如果 *y 是引用

*y 是不是参考.y 是一个参考；*y dereferences y，允许您访问被引用的值.

*y is not a reference. y is a reference; *y dereferences y, allowing you access to the referred-to value.

[+= 和 println! 之间有什么区别]

what is the difference [between += and println!]

println! 是一个自动引用的宏给它的参数.此外，Display trait(通过格式字符串中的 {} 使用)是为所有对自身实现 Display 的类型的引用实现的(impl<'a, T> Display for &'a T where T: Display + ?Sized).

println! is a macro that automatically references the arguments given to it. In addition, the Display trait (used via {} in the format string) is implemented for all references to types that themselves implement Display (impl<'a, T> Display for &'a T where T: Display + ?Sized).

因此，println!("{}", y); 实际上是打印出对值引用的引用.由于 Display 的实现，这些中间引用会自动取消引用.

Thus, println!("{}", y); is actually printing out a reference to a reference to a value. Those intermediate references are automatically dereferenced due to the implementation of Display.

+= 是通过 AddAssign 特性.标准库仅实现向自身添加整数类型(impl AddAssign for i32).这意味着您必须添加适当级别的取消引用才能使两边都为整数.

+=, on the other hand, is implemented via the AddAssign trait. The standard library only implements adding an integer type to itself (impl AddAssign<i32> for i32). That means that you have to add an appropriate level of dereferencing in order to get both sides to an integer.

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