如何在没有记忆泄漏的情况下捕获铁锈信号

Question

我注意到程序中的内存泄漏，并将其跟踪到信号处理。似乎没有一种无泄漏的方法来做到这一点。我不担心Valgrind报告的“仍然可以达到的”字节 - 我担心“可能丢失”字节。

最小可重复的示例：

use tokio::signal;
use tokio::time::{sleep, Duration};

async fn sleep1() {
    loop {
        sleep(Duration::from_secs(1)).await;
        println!("sleep1");
    }
}

async fn sleep2() {
    loop {
        sleep(Duration::from_secs(2)).await;
        println!("sleep2");
    }
}

#[tokio::main]
async fn main() -> Result<(), Box<dyn std::error::Error>> {
    let s1 = Box::pin(sleep1());
    let s2 = Box::pin(sleep2());
    let sig = Box::pin(signal::ctrl_c());
    tokio::select! {
        _ = s1 => {},
        _ = s2 => {},
        _ = sig => {},
    };
    println!("shutting down");
    Ok(())
}

摘录摘自货物。汤姆尔文件：

edition = "2021"
tokio = { version = "1", features = ["full"] }

valgrind输出：

==1366460== Command: target/debug/simulation
==1366460== 
sleep1
sleep2
sleep1
sleep1
^Cshutting down
==1366460== 
==1366460== HEAP SUMMARY:
==1366460==     in use at exit: 25,884 bytes in 82 blocks
==1366460==   total heap usage: 617 allocs, 535 frees, 145,635 bytes allocated
==1366460== 
==1366460== LEAK SUMMARY:
==1366460==    definitely lost: 0 bytes in 0 blocks
==1366460==    indirectly lost: 0 bytes in 0 blocks
==1366460==      possibly lost: 1,188 bytes in 3 blocks
==1366460==    still reachable: 24,696 bytes in 79 blocks
==1366460==         suppressed: 0 bytes in 0 blocks
==1366460== Rerun with --leak-check=full to see details of leaked memory
==1366460== 
==1366460== For lists of detected and suppressed errors, rerun with: -s
==1366460== ERROR SUMMARY: 0 errors from 0 contexts (suppressed: 0 from 0)

Answer 1

根据Tokio开发人员的说法，这是使用全局变量的假阳性。参见在这里。

另外，我发现，如果一个人真的希望摆脱valgrind错误，则有可能 and 从rust 称呼它。 RUST程序将产生一个阻止线程（例如Tokio :: tos :: Spawn_blocking），该线程等待C代码捕获信号并终止。