feat(move_semantics2): rewrite hint

info.toml

@@ -287,23 +287,24 @@ Also: Try accessing `vec0` after having called `fill_vec()`. See what happens!""
 [[exercises]]
 name = "move_semantics2"
 path = "exercises/move_semantics/move_semantics2.rs"
-mode = "compile"
+mode = "test"
 hint = """
-So, `vec0` is passed into the `fill_vec` function as an argument. In Rust,
-when an argument is passed to a function and it's not explicitly returned,
-you can't use the original variable anymore. We call this "moving" a variable.
-Variables that are moved into a function (or block scope) and aren't explicitly
-returned get "dropped" at the end of that function. This is also what happens here.
-There's a few ways to fix this, try them all if you want:
-1. Make another, separate version of the data that's in `vec0` and pass that
+When running this exercise for the first time, you'll notice an error about
+"borrow of moved value". In Rust, when an argument is passed to a function and
+it's not explicitly returned, you can't use the original variable anymore.
+We call this "moving" a variable. When we pass `vec0` into `fill_vec`, it's being
+"moved" into `vec1`, meaning we can't access `vec0` anymore after the fact.
+Rust provides a couple of different ways to mitigate this issue, feel free to try them all:
+1. You could make another, separate version of the data that's in `vec0` and pass that
    to `fill_vec` instead.
 2. Make `fill_vec` borrow its argument instead of taking ownership of it,
-   and then copy the data within the function in order to return an owned
-   `Vec<i32>`
-3. Make `fill_vec` *mutably* borrow a reference to its argument (which will need to be
-   mutable), modify it directly, then not return anything. Then you can get rid
-   of `vec1` entirely -- note that this will change what gets printed by the
-   first `println!`"""
+   and then copy the data within the function (`vec.clone()`) in order to return an owned
+   `Vec<i32>`.
+3. Or, you could make `fill_vec` *mutably* borrow a reference to its argument (which will need to be
+   mutable), modify it directly, then not return anything. This means that `vec0` will change over the
+   course of the function, and makes `vec1` redundant (make sure to change the parameters of the `println!`
+   statements if you go this route)
+"""
 
 [[exercises]]
 name = "move_semantics3"