Why do I get "Borrowed value does not live long enough" in this example?

Question

Editor's note: The code in this question predates Rust 1.0. The equivalent modern version of this code compiles as-is.

I'm still taking baby steps at learning Rust, and was surprised at the following.

I couldn't understand why this code compiles:

use std::iter::AdditiveIterator;

fn main() {

    let range = [1,2,3,4,5,6,7,8,9];
    let sum = range.iter().map(|&x| x * x).filter(|&x| x % 2 == 0).sum();

    println!("{}", sum);
}

While this doesn't: (just moving the .iter() up)

use std::iter::AdditiveIterator;

fn main() {

    let range = [1,2,3,4,5,6,7,8,9].iter();
    let sum = range.map(|&x| x * x).filter(|&x| x % 2 == 0).sum();

    println!("{}", sum);
}

Which renders this error:

test.rs:5:17: 5:36 error: borrowed value does not live long enough
test.rs:5     let range = [1,2,3,4,5,6,7,8,9].iter();
                          ^~~~~~~~~~~~~~~~~~~

I'm sure it has something to do with Rust scopes etc, but I'm not sure I understand how by just moving the method call to a different line makes a difference.

Answer 1

In modern Rust, the equivalent code compiles:

fn main() {
    let range = [1, 2, 3, 4, 5, 6, 7, 8, 9].iter();
    let sum: i32 = range.map(|&x| x * x).filter(|&x| x % 2 == 0).sum();
    println!("{}", sum);
}

This is because literals, like the array literal, are now automatically promoted to static variables if they need to be. The result of iter here is now a std::slice::Iter<'static, i32>.

See also:

• Why can I return a reference to a local literal but not a variable?

Answer 2

The array gets destroyed at the end of the let range = [1,2,3,4,5,6,7,8,9].iter(); statement as there is no variable that holds that vector. This would cause a dangling iterator pointing nowhere.

The same occurs in C++, where it is possible to create iterator on object that gets deleted after.