将可变指针传递给函数而不先初始化它

Question

use std::ptr::{addr_of_mut, null_mut};

use libc::{CLOCK_MONOTONIC, timer_create, timer_delete, timer_t};

fn main() {
    let mut timer1: timer_t = null_mut();

    unsafe {
        let r = timer_create(CLOCK_MONOTONIC, null_mut(), addr_of_mut!(timer1));
        if r == 0 {
            timer_delete(timer1);
        }
    }
}

调用timer_create()时，生成的计时器ID存储在变量timer1中。我将它作为可变指针传递，所以这是输出变量。

我知道 API 保证它是安全的，如何避免像上面的代码那样将 timer1 初始化为 null_mut()？

Answer 1

您可以使用 MaybeUninit:

use std::mem::MaybeUninit;
use std::ptr::null_mut;

use libc::{CLOCK_MONOTONIC, timer_create, timer_delete, timer_t};

fn main() {
    let mut timer1 = MaybeUninit::<timer_t>::uninit();
    let timer1 = unsafe {
        let r = timer_create(CLOCK_MONOTONIC, null_mut(), timer1.as_mut_ptr());
        if r != 0 {
            panic!("…");
        }
        timer1.assume_init()
    };
    unsafe {
        timer_delete(timer1);
    }
}

将可变指针传递给函数而不先初始化它

Pass mutable pointer to function without initializing it first

ffi

rust