将新构造的数组传递给函数时出现意外的生命周期问题

Question

我一直在开发一个旨在在 Solana 运行时上运行的程序。我最近遇到了一个问题，数组中的第二项被传递到“test2”中会出现某种生命周期错误。当显式生命周期注释添加到 test1 时，此错误消失，但我无法弄清楚它为什么首先出现。

pub fn test1(
    test_account_1: &AccountInfo,
    test_account_2: &AccountInfo,
    mint_account: &AccountInfo,
) -> ProgramResult {
    test2(&[test_account_1.clone(), test_account_2.clone()]);
    return Ok(());
}

pub fn test2(arr: &[AccountInfo]) {
    for a in arr.iter() {
        println!("{}", a.key)
    }
}

错误：

寿命不匹配

...但是来自test_account_1 的数据流入test_account_2 这里是 rustc (E0623)

lib.rs(207, 22)：这两种类型声明为 不同的生命周期...

lib.rs(208, 22):

lib.rs(211, 37): ...但是数据 从test_account_1 流入test_account_2 这里

AccountInfo 的定义：

/// Account information
#[derive(Clone)]
pub struct AccountInfo<'a> {
    /// Public key of the account
    pub key: &'a Pubkey,
    /// Was the transaction signed by this account's public key?
    pub is_signer: bool,
    /// Is the account writable?
    pub is_writable: bool,
    /// The lamports in the account.  Modifiable by programs.
    pub lamports: Rc<RefCell<&'a mut u64>>,
    /// The data held in this account.  Modifiable by programs.
    pub data: Rc<RefCell<&'a mut [u8]>>,
    /// Program that owns this account
    pub owner: &'a Pubkey,
    /// This account's data contains a loaded program (and is now read-only)
    pub executable: bool,
    /// The epoch at which this account will next owe rent
    pub rent_epoch: Epoch,
}

Answer 1

为了让您的数组包含对那些AccountsInfo 的引用，它们需要至少存在相同的生命周期。由于它们具有内部边界，因此它们也需要匹配：

pub fn test1<'a>(
    test_account_1: &AccountInfo<'a>,
    test_account_2: &AccountInfo<'a>,
    mint_account: &AccountInfo,
) -> Result<(), ()> {
    test2(&[test_account_1.clone(), test_account_2.clone()]);
    Ok(())
}

Playground

另外，根据您的实际需要，您可能根本不需要克隆：

pub fn test1<'a>(
    test_account_1: &AccountInfo<'a>,
    test_account_2: &AccountInfo<'a>,
    mint_account: &AccountInfo,
) -> Result<(), ()> {
    test2(&[test_account_1, test_account_2]);
    Ok(())
}

pub fn test2(arr: &[&AccountInfo]) {
    for a in arr.iter() {
        println!("{}", a.key)
    }
}

Playground