为什么打印指针与打印取消引用的指针打印的内容相同? [英] Why does printing a pointer print the same thing as printing the dereferenced pointer?

问题描述

来自 Rust 指南:

From the Rust guide:

要取消引用(获取被引用的值而不是引用本身)y，我们使用星号 (*)

To dereference (get the value being referred to rather than the reference itself) y, we use the asterisk (*)

所以我做到了:

fn main() {
    let x = 1;
    let ptr_y = &x;
    println!("x: {}, ptr_y: {}", x, *ptr_y);
}

即使没有显式取消引用，这也给了我相同的结果 (x=1; y=1):

This gives me the same results (x=1; y=1) even without an explicit dereference:

fn main() {
    let x = 1;
    let ptr_y = &x;
    println!("x: {}, ptr_y: {}", x, ptr_y);
}

为什么?ptr_y 不应该打印内存地址而 *ptr_y 打印 1 吗?是否有某种自动取消引用或我错过了什么?

Why? Shouldn't ptr_y print the memory address and *ptr_y print 1? Is there some kind of auto-dereference or did I miss something?

推荐答案

Rust 通常关注对象值(即内容中有趣的部分)而不是对象标识(内存地址).Display for &T 其中 T 实现 Display 直接遵循内容.为 Display 的 String 实现手动扩展该宏:

Rust usually focuses on object value (i.e. the interesting part of the contents) rather than object identity (memory addresses). The implementation of Display for &T where T implements Display defers directly to the contents. Expanding that macro manually for the String implementation of Display:

impl<'a> Display for &'a String {
    fn fmt(&self, f: &mut Formatter) -> Result {
        Display::fmt(&**self, f)
    }
}

也就是说，它只是直接打印其内容.

That is, it is just printing its contents directly.

如果你关心对象身份/内存地址，你可以使用Pointer 格式化程序, {:p}:

If you care about object identity/the memory address, you can use the Pointer formatter, {:p}:

fn main() {
    let x = 1;
    let ptr_y = &x;
    println!("x: {}, ptr_y: {}, address: {:p}", x, ptr_y, ptr_y);
}

输出:

x: 1, ptr_y: 1, address: 0x7fff4eda6a24

游乐场

这篇关于为什么打印指针与打印取消引用的指针打印的内容相同?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！