In Rust, memory is managed through a system of ownership with a set of rules that the compiler checks at compile time.

Rust follows three fundamental rules of ownership.

• Each value in rust has a variable that's called its owner.

let x = String::from("hello"); // string "hello" has its owner: x

• There can be only one owner at a time.

let x = String::from("hello");
let text = x; // string "hello" has been moved to text.
println!("x value is {x}"); // show compile error. x's value has been moved out.

• When the owner goes out of scope, the value will be dropped.

fn my_func() {
	let x = String::from("hello");
	println!("x value is {x}");
} // x is out of scope, the value "hello" will be dropeed.

The ownership of variable (except from primitive types) follows the same pattern every time: assigning a value to another variable moves it.

Stack Only Data : Copy

All primitive types implement the Copy trait and their ownership is therefore not moved like one would assume, following the ‘ownership rules’. ^[1]

• Primitive data are known-sized data at compile time, so rust copy the variable instead of move.

let x = 10;
let y = x; // y is copied from x
println!("x is {x} and y is {y}");

compound type as example:

let arr: [i32; 5] = [1,2,3,4,5];
let arr_2 = arr; // copy on assignment.
println!("arr is {:} and arr_2 is {:}", arr, arr_2);

Exceptions

Any primitive compound type that contains non-primitive types (e.g. String or Vector) are still follow move on assignment.

arr_2 is move on assignment rather than copy.

let arr = [String::from("Chester"), String::from("John"), String::from("Marry")];
let arr_2 = arr; // move on assignment

tuple_2 is move on assignment rather than copy.

let tuple = (100, String::from("hello"));
let tuple_2 = tuple; // move on assignment

Moves on Index Content

We can not move out the index of Vector, e.g.

let e = v[1] is forbidden because we can not move out the indexed content from the owned variable.

let v = vec!["1".to_string(), "2".to_string(), "3".to_string()];
let e = v[1];

How to fix it ?
Use reference:

let v = vec!["1".to_string(), "2".to_string(), "3".to_string()];
let e = &v[0];