rust - 为什么对象的地址会在方法之间发生变化？

Question

我有要连接的 C 代码。我选择使用 mem::uninitialized 先声明内存，然后调用 C 函数 (UserInit) 进行初始化，然后使用它(在 UserDoSomething)。

奇怪的是对象的地址在 UserInit 和 UserDoSomething 中是不同的。为什么它会这样？

C 代码:

typedef struct {
    char* name;
    int32_t age;
} User;

void
UserInit(User* u){
    printf("in init: user addr: %p\n", u);
}

void
UserDoSomething(User* u){
    printf("in do something user addr: %p\n", u);
}

void
UserDestroy(User* u){
    free(u->name);
}

Rust FFI:

use std::mem;
use std::os::raw::c_char;
use std::ffi::CString;


#[repr(C)]
pub struct User{
    pub name:   *const c_char,
    pub age:    i32,
}

impl User {
    pub fn new()-> User {

        let ret: User = unsafe { mem::uninitialized() };

        unsafe {
            UserInit(&mut ret as *mut User)
        }

        ret
    }

    pub fn do_something(&mut self){
        unsafe {
            UserDoSomething(self as *mut User)
        }
    }

}
extern "C" {
    pub fn UserInit(u:*mut User);
    pub fn UserDoSomething(u:*mut User);
    pub fn UserDestroy(u:*mut User);
}

使用rust 测试:

mod ffi;

use ffi::User;

fn main() {
    let mut u = User::new();
    u.do_something();
}

理论上，它应该输出相同的地址，但事实并非如此:

> cargo run
     Running `target/debug/learn`
in init: user addr: 0x7fff5b948b80
in do something user addr: 0x7fff5b948ba0

Answer 1

That's just how Rust works.

不仅如此，C 的工作方式也是如此:

#include <stdint.h>
#include <stdlib.h>
#include <stdio.h>

typedef struct {
    char* name;
    int32_t age;
} User;

void
UserInit(User* u){
    printf("in init: user addr: %p\n", u);
}

void
UserDoSomething(User* u){
    printf("in do something user addr: %p\n", u);
}

void
UserDestroy(User* u){
    free(u->name);
}

User rust_like_new(void) {
    User u;
    UserInit(&u);
    return u;
}

int main(int argc, char *argv[]) {
    User u = rust_like_new();
    UserDoSomething(&u);
}

in init: user addr:        0x7fff506c1600
in do something user addr: 0x7fff506c1630

通常，您不关心容器的地址，只关心它包含什么。

If I heap allocate User, the address won't change, but if I use Box (let u = Box::new(User::new())), it still changes.

同样的事情发生在 Rust 和 C 中。Box<User> 的地址或 User *本身就会改变。 Box<User> 的值(指向的东西)或 User *将保持一致。

mod ffi {
    use std::mem;
    use std::os::raw::c_char;

    #[repr(C)]
    pub struct User {
        pub name: *const c_char,
        pub age: i32,
    }

    impl User {
        pub fn new() -> Box<User> {
            let mut ret: Box<User> = Box::new(unsafe { mem::uninitialized() });

            unsafe { UserInit(&mut *ret) }

            ret
        }

        pub fn do_something(&mut self) {
            unsafe { UserDoSomething(self) }
        }
    }

    extern "C" {
        pub fn UserInit(u: *mut User);
        pub fn UserDoSomething(u: *mut User);
    }
}

use ffi::User;

fn main() {
    let mut u = User::new();
    u.do_something();
}

in init: user addr:        0x10da17000
in do something user addr: 0x10da17000

如果您传递对 User 的引用到 C before 它被移动到 Box , 那么是的，地址将在移入 Box 时发生变化.这相当于:

User rust_like_new(void) {
    User u;
    UserInit(&u);
    return u;
}

int main(int argc, char *argv[]) {
    User u = rust_like_new();
    User *u2 = malloc(sizeof(User));
    *u2 = u;
    UserDoSomething(u2);
}

请注意，Rust(和其他语言)允许 RVO被执行。但是，我认为打印出地址 会取消此优化的资格，因为如果启用 RVO，行为会发生变化。您需要查看调试器或生成的程序集。