rust - 从函数返回 future 值

Question

我最近开始学习Rust，但不确定如何从应该返回Result的函数中返回将来的值。当我尝试仅返回响应变量并删除Result输出时，出现错误:无法在返回? 的函数中使用std::string::String运算符

#[tokio::main]
async fn download() -> Result<(),reqwest::Error> {
    let url = "https://query1.finance.yahoo.com/v8/finance/chart/TSLA";
    let response = reqwest::get(url)
                            .await?
                            .text()
                            .await?;
    Ok(response)
 }

我对main()的期望是获取并打印响应值:

fn main() {
    let response = download();
    println!("{:?}", response)
}

Answer 1

我想你的代码应该看起来像这样

extern crate tokio; // 0.2.13

async fn download() -> Result<String, reqwest::Error> {
    let url = "https://query1.finance.yahoo.com/v8/finance/chart/TSLA";

    reqwest::get(url).await?.text().await
}

#[tokio::main]
async fn main() {
    let response = download().await;

    println!("{:?}", response)
}

这是rust playground link