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Why can I return a reference to a local literal but not a variable?（1个答案）
11天前关闭。
我正在Rust中学习引用和切片，所以尝试写一些随机代码：

fn main() {
    let a = &[&[2]][..];
    println!("{:?}", a); // OK

    let b = 2;
    let c = &[&[b]][..];
    println!("{:?}", c);
//     error[E0716]: temporary value dropped while borrowed
//  --> src/main.rs:6:16
//   |
// 6 |     let c = &[&[b]][..];
//   |                ^^^     - temporary value is freed at the end of this statement
//   |                |
//   |                creates a temporary which is freed while still in use
// 7 |     println!("{:?}", c);
//   |                      - borrow later used here
//   |
// help: consider using a `let` binding to create a longer lived value
//   |
// 6 ~     let binding = [b];
// 7 ~     let c = &[&binding][..];
//   |

// For more information about this error, try `rustc --explain E0716`.
// error: could not compile `playground` due to previous error
}

我不知道为什么把&[2]改成&[b]会出现上面的错误，请你给我解释一下。
作为一个新手，我很难把我的问题解释得更清楚。对不起，伙计们!!