What does "cannot move out of index of" mean? - rust

Question

I am playing with Rust, and I'm trying to access the first command line argument with this code:
use std::env;
fn main() {
let args: Vec<_> = env::args().collect();
let dir = args[1];
}
And I get this error:
error[E0507]: cannot move out of indexed content
--> src/main.rs:5:15
|
5 | let dir = args[1];
| --- ^^^^^^^ cannot move out of indexed content
| |
| hint: to prevent move, use `ref dir` or `ref mut dir`
Or in later versions of Rust:
error[E0507]: cannot move out of index of `std::vec::Vec<std::string::String>`
--> src/main.rs:5:15
|
5 | let dir = args[1];
| ^^^^^^^
| |
| move occurs because value has type `std::string::String`, which does not implement the `Copy` trait
| help: consider borrowing here: `&args[1]`
If I change it to let ref dir, it compiles, but I don't grok what's going on. Could someone explain what "indexed content" means?

Answer 1

When you use an index operator ([]) you get the actual object at index location. You do not get a reference, pointer or copy. Since you try to bind that object with a let binding, Rust immediately tries to move (or copy, if the Copy trait is implemented).
In your example, env::args() is an iterator of Strings which is then collected into a Vec<String>. This is an owned vector of owned strings, and owned strings are not automatically copyable.
You can use a let ref binding, but the more idiomatic alternative is to take a reference to the indexed object (note the & symbol):
use std::env;
fn main() {
let args: Vec<_> = env::args().collect();
let ref dir = &args[1];
// ^
}
Implicitly moving out of a Vec is not allowed as it would leave it in an invalid state — one element is moved out, the others are not. If you have a mutable Vec, you can use a method like Vec::remove to take a single value out:
use std::env;
fn main() {
let mut args: Vec<_> = env::args().collect();
let dir = args.remove(1);
}
See also:
What is the return type of the indexing operation?
For your particular problem, you can also just use Iterator::nth:
use std::env;
fn main() {
let dir = env::args().nth(1).expect("Missing argument");
}

Answer 2

The accepted answer has already given the solution. I would like to explain this problem on a semantic level as a complement.
The rule is: A borrowed value can't be moved out. See this: E0507
[] operator came from the Index trait, whose function signature is:
fn index(&self, index: I) -> &<Vec<T, A> as Index<I>>::Output
As you can see, it return a reference, not own the value. Moving it out break the rule mentioned above.