Why do not work use a variable after another variable with the ownership be dropped?

[firmino]

fn main() {
    let a = vec![2, 4, 8];
    {
        let b = a; // move value of 'a' to 'b'
        println!("b : {:?} ", b); // prints 'b'
    }
    // println!("{}", b) // Error, b is out of scope (dropped)
    println!("a : {:?} ", a); // Error; borrow of moved value  <<<<<< WHY CAN'T DO THIS? The ownership didn't back to a since b is dropped?
}

[Hyeonu]

If it's allowed it would be an use-after-free bug. The type of the variable b is Vec<i32> so it will drop its contents and buffer when it goes out of its scope.

It's borrowing which ends automatically when it goes out of scope. Ownership never come back implicitly.

[jdahlstrom]

Ownership never ”goes back” automatically. When b is dropped, whatever resources it owns are freed and can’t even in principle go back to a previous owner, if any. If you give me a lollipop, as a gift, and then I eat it, you can’t expect to get it back.

What you need instead is to borrow, ie. to use a reference. Borrowing is temporary and the original gets to use a borrowed value again once the borrow goes out of scope.

[kornel]

Owned values can only exist in one place at a time. The vec can't be in both a and b.

It works if you move the value back to a explicitly:

fn main() {
    let mut a = vec![2, 4, 8];
    {
        let b = a; // move value of 'a' to 'b'
        println!("b : {:?} ", b); // prints 'b'
        a = b; // give it back
    }
    println!("a : {:?} ", a); 
}

[firmino]

Thank you very much for the answer. I got it! Perfect example. For me, a=b and a=& b would have the same behavior, but I understood that is not the same.