小编Nig*_*eXD的帖子

我有一个下载异步文件的函数：

async fn download_file() {
    fn main() {
        let resp = reqwest::blocking::get("https://sh.rustup.rs").expect("request failed");
        let body = resp.text().expect("body invalid");
        let mut out = File::create("rustup-init.sh").expect("failed to create file");
        io::copy(&mut body.as_bytes(), &mut out).expect("failed to copy content");
    }
}

我想调用这个函数来下载文件，然后在需要时等待它。
但问题是，如果我这样做，我会得到一个错误：

fn main() {
    let download = download_file();
    // Do some work
    download.await; // `await` is only allowed inside `async` functions and blocks\nonly allowed inside `async` functions and blocks
    // Do some work
}

所以我必须使 main 函数异步，但是当我这样做时，我收到另一个错误：

async fn main() { // `main` function …

i我想指定这个 for 循环中的类型。Visual Studio Code 告诉我输入的类型，i32但数字不是那么大，所以我想将其更改为u8.

for i in 0..101 {
    println!("{}", i);
}

我试过这个：

for i: u8 in 0..101 {
    println!("{}", i);
}

但是，我收到此错误：

error: expected one of `@` or `|`, found `:`
 --> src/main.rs:2:10
  |
2 |     for i: u8 in 0..101 {
  |          ^
  |          |
  |          expected one of `@` or `|`
  |          help: maybe write a path separator here: `::`

指定类型的正确方法是什么？

我在标准库中查看了String很多unsafe这样的代码：

    #[inline]
    #[stable(feature = "rust1", since = "1.0.0")]
    pub fn remove(&mut self, idx: usize) -> char {
        let ch = match self[idx..].chars().next() {
            Some(ch) => ch,
            None => panic!("cannot remove a char from the end of a string"),
        };

        let next = idx + ch.len_utf8();
        let len = self.len();
        unsafe {
            ptr::copy(self.vec.as_ptr().add(next), self.vec.as_mut_ptr().add(idx), len - next);
            self.vec.set_len(len - (next - idx));
        }
        ch
    }

unsafe为什么标准库里有这么多代码？语言如何仍然安全？

首先，我认为编译时间会永远持续下去，或者我遇到一个奇怪的错误，但这并没有发生。代码运行了一段时间然后崩溃了。

这是我的代码：

#include <iostream>

inline void say_hello()
{
    std::cout << "hello\n";
    say_hello();
}

int main()
{
    say_hello();
}

我以为编译器会把它转换成这样的：

#include <iostream>

int main()
{
    std::cout << "hello\n";
    std::cout << "hello\n";
    std::cout << "hello\n";
    std::cout << "hello\n";
    std::cout << "hello\n";
    std::cout << "hello\n";
    std::cout << "hello\n";
    std::cout << "hello\n";
    std::cout << "hello\n";
    std::cout << "hello\n";
    std::cout << "hello\n";
    std::cout << "hello\n";
    std::cout << "hello\n";
    std::cout << "hello\n";
    std::cout << "hello\n";
    // and crash because my storage is filled
}

但是，我认为编译器忽略了该inline …