Pass mutable pointer to function without initializing it first

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);
        }
    }
}

When calling timer_create(), the resulting timer ID is stored at variable timer1. I pass it as a mutable pointer, so that's the output variable.

How can I avoid initializing timer1 to null_mut() as in the code above knowing that it's guaranteed by the API to be safe ?

Answer 1

You can use 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);
    }
}